Question 1

Consider the following balanced reaction.What mass (in g)of CO₂ can be formed from 288 mg of O₂? Assume that there is excess C₃H₇SH present. C₃H₇SH(l)+ 6 O₂(g)→ 3 CO₂(g)+ SO₂(g)+ 4 H₂O(g)

Accepted Answer

E

Question 2

How many molecules of H₂S are required to form 79.0 g of sulfur according to the following reaction? Assume excess SO₂. 2 H₂S(g)+ SO₂(g)→ 3 S(s)+ 2H₂O(l)

Accepted Answer

First, we need to find the moles of sulfur using its molar mass: $$79.0 ext{ g S} imes \frac{1 ext{ mol S}}{32.066 ext{ g S}} = 2.46 ext{ mol S}$$ According to the balanced equation, 2 mol of H₂S produce 3 mol of S, so we can set up a proportion to find the moles of H₂S: $$\frac{2 ext{ mol H₂S}}{3 ext{ mol S}} = \frac{x ext{ mol H₂S}}{2.46 ext{ mol S}}$$ Solving for x, we get: $$x = \frac{2.46 ext{ mol S} imes 2 ext{ mol H₂S}}{3 ext{ mol S}} = 1.64 ext{ mol H₂S}$$ Finally, we can convert mol H₂S to molecules of H₂S using Avogadro's number: $$1.64 ext{ mol H₂S} imes 6.022 imes 10^{23} ext{ molecules/mol} = 9.89 imes 10^{23} ext{ molecules of H₂S}$$ Therefore, the answer is choice B.

Question 3

Determine the theoretical yield of HCl if 60.0 g of BCl₃and 37.5 g of H₂O are reacted according to the following balanced reaction.A possibly useful molar mass is BCl₃ = 117.16 g/mol. BCl₃(g)+ 3 H₂O(l)→ H₃BO₃(s)+ 3 HCl(g)

Accepted Answer

First, calculate the moles of BCl3 and H2O. For BCl3: $60.0 \, \text{g} \div 117.16 \, \text{g/mol} = 0.512 \, \text{mol}$. For H2O: $37.5 \, \text{g} \div 18.015 \, \text{g/mol} = 2.08 \, \text{mol}$. The reaction shows BCl3 reacts with 3 moles of H2O to produce 3 moles of HCl. BCl3 is the limiting reagent, so it will produce $0.512 \, \text{mol} \times 3 = 1.536 \, \text{mol}$ of HCl. The mass of HCl produced is $1.536 \, \text{mol} \times 36.46 \, \text{g/mol} = 56.0 \, \text{g}$.

Question 4

Consider the following reaction.How many moles of oxygen are required to produce 2.33 moles of water? Assume that there is excess C₃H₇SH present. C₃H₇SH(l)+ 6 O₂(g)→ 3 CO₂(g)+ SO₂(g)+ 4 H₂O(g)

Accepted Answer

The answer of Consider the following reaction.How many moles of...

Question 5

A 12.39 g sample of phosphorus reacts with 42.54 g of chlorine to form only phosphorus trichloride (PCl₃).If it is the only product,what mass of PCl₃ is formed?

Accepted Answer

First, we need to balance the equation:

P + 3Cl2 → PCl3

Next, we need to determine the limiting reagent. To do this, we can calculate the amount of PCl3 that would be formed if all of the P and Cl2 reacted completely:

Using P:

12.39 g P × (1 mol P/30.97 g P) × (1 mol PCl3/1 mol P) × (137.33 g PCl3/1 mol PCl3) = 17.01 g PCl3

Using Cl2:

42.54 g Cl2 × (1 mol Cl2/70.90 g Cl2) × (1 mol PCl3/3 mol Cl2) × (137.33 g PCl3/1 mol PCl3) = 86.92 g PCl3

Since we have less P than Cl2, P is the limiting reagent. Therefore, the amount of PCl3 formed will be determined by the amount of P:

12.39 g P × (1 mol P/30.97 g P) × (1 mol PCl3/1 mol P) × (137.33 g PCl3/1 mol PCl3) = 17.01 g PCl3

Therefore, the answer is B) 54.93 g.

Question 6

Determine the limiting reactant (LR)and the mass (in g)of nitrogen that can be formed from 50.0 g N₂O₄ and 45.0 g N₂H₄.Some possibly useful molar masses are as follows: N₂O₄ = 92.02 g/mol,N₂H₄ = 32.05 g/mol. N₂O₄(l)+ 2 N₂H₄(l)→ 3 N₂(g)+ 4 H₂O(g)

Accepted Answer

To find the limiting reactant, we calculate the moles of each reactant and then use the stoichiometry of the reaction. For $N_2O$ (92.02 g/mol), $50.0 \, \text{g} / 92.02 \, \text{g/mol} = 0.543 \, \text{moles}$. For $N_2H_4$ (32.05 g/mol), $45.0 \, \text{g} / 32.05 \, \text{g/mol} = 1.404 \, \text{moles}$. The reaction requires 2 moles of $N_2H_4$ for every mole of $N_2O$, so $N_2O$ is the limiting reactant because the ratio of $N_2H_4$ to $N_2O$ is less than 2:1. From the stoichiometry, 1 mole of $N_2O$ produces 3 moles of $N_2$, so $0.543 \, \text{moles} \times 3 = 1.629 \, \text{moles of } N_2$. The molar mass of $N_2$ is $14.01 \times 2 = 28.02 \, \text{g/mol}$, so $1.629 \, \text{moles} \times 28.02 \, \text{g/mol} = 45.7 \, \text{g}$.

Question 7

How many grams of Li₃N can be formed from 1.75 moles of Li? Assume an excess of nitrogen. 6 Li(s)+ N₂(g)→ 2 Li₃N(s)

Accepted Answer

The answer of How many grams of Li₃N can be...

Question 8

How many molecules of HCl are formed when 50.0 g of water reacts according to the following balanced reaction? Assume excess ICl₃. 2 ICl₃ + 3 H₂O → ICl + HIO₃ + 5 HCl

Accepted Answer

50.0 g of water is 50.0 g / 18.02 g/mol = 2.77 mol of water. From the balanced reaction, we see that 3 mol of water react to produce 5 mol of HCl. Therefore, 2.77 mol of water will react to produce 2.77 mol * 5 mol HCl / 3 mol H2O = 4.62 mol of HCl. 4.62 mol of HCl is 4.62 mol * 6.022 * 10^23 molecules/mol = 2.78 * 10^24 molecules of HCl.

Question 9

Consider the following balanced reaction.How many grams of water are required to form 75.9 g of HNO₃? Assume that there is excess NO₂ present.The molar masses are as follows: H₂O = 18.02 g/mol,HNO₃ = 63.02 g/mol. 3 NO₂(g)+ H₂O(l)→ 2 HNO₃(aq)+ NO(g)

Accepted Answer

The answer of Consider the following balanced reaction.How many grams...

Question 10

How many grams of oxygen are formed when 6.21 moles of KOH are formed? 4 KO(s)+ 2 H₂O(l)→ 4 KOH(s)+ O₂(g)

Accepted Answer

The answer of How many grams of oxygen are formed...

Question 11

How many moles of oxygen are formed when 58.6 g of KNO₃ decomposes according to the following reaction? The molar mass of KNO₃ is 101.11 g/mol. 4 KNO₃(s)→ 2 K₂O(s)+ 2 N₂(g)+ 5 O₂(g)

Accepted Answer

First, we need to find the number of moles of KNO₃: 58.6 g KNO₃ / 101.11 g/mol KNO₃ = 0.5796 mol KNO₃ According to the balanced equation, for every 4 moles of KNO₃, 5 moles of O₂ are produced. Therefore, we can set up a proportion: 4 mol KNO₃ / 5 mol O₂ = 0.5796 mol KNO₃ / x mol O₂ Solving for x, we get: x mol O₂ = (5 mol O₂)(0.5796 mol KNO₃) / (4 mol KNO₃) = 0.7245 mol O₂ Rounding to three significant figures, we get: x mol O₂ = 0.724 mol O₂ Therefore, the answer is D) 0.724 mol O₂.

Question 12

According to the following balanced reaction,how many moles of NO are formed from 8.44 moles of NO₂if there is plenty of water present? 3 NO₂(g)+ H₂O(l)→ 2 HNO₃(aq)+ NO(g)

Accepted Answer

According to the balanced reaction, 3 moles of NO2 react to form 1 mole of NO. Therefore, 8.44 moles of NO2 will react to form 8.44/3 = 2.81 moles of NO.

Question 13

According to the following reaction,how many grams of sulfur are formed when 37.4 g of water are formed? 2 H₂S(g)+ SO₂(g)→ 3 S(s)+ 2H₂O(l)

Accepted Answer

The answer of According to the following reaction,how many grams...

Question 14

A 14.01 g sample of N₂ reacts with 3.02 g of H₂ to form ammonia (NH₃).If ammonia is the only product,what mass of ammonia is formed?

Accepted Answer

The mass of ammonia formed is the sum of the masses of the reactants (14.01 g + 3.02 g = 17.03 g), which is closest to 17.01 g, due to the law of conservation of mass.

Question 15

According to the following balanced reaction,how many moles of KO are required to exactly react with 5.44 moles of H₂O? 4 KO(s)+ 2 H₂O(l)→ 4 KOH(s)+ O₂(g)

Accepted Answer

According to the balanced reaction, it takes 2 moles of H₂O to react with 4 moles of KO. So, the molar ratio of H₂O to KO is 2:4 or 1:2. Therefore, to find the number of moles of KO required to react with 5.44 moles of H₂O, we need to divide 5.44 by 2: 5.44 moles H₂O ÷ 2 = 2.72 moles KO Therefore, the correct answer is C, 10.9 moles H₂O.

Question 16

Carbonic acid can form water and carbon dioxide upon heating.How much carbon dioxide is formed from 6.20 g of carbonic acid? H₂CO₃ → H₂O + CO₂

Accepted Answer

The molar mass of carbonic acid (H2CO3) is 62.01 g/mol, and the molar mass of carbon dioxide (CO2) is 44.01 g/mol. Therefore, 6.20 g of carbonic acid is equal to 6.20 g / 62.01 g/mol = 0.1 mol of carbonic acid. According to the balanced chemical equation, 1 mol of carbonic acid produces 1 mol of carbon dioxide. Therefore, 0.1 mol of carbonic acid will produce 0.1 mol of carbon dioxide, which is equal to 0.1 mol * 44.01 g/mol = 4.40 g of carbon dioxide.

Question 17

According to the following reaction,what amount of Al₂S₃ remains when 20.00 g of Al₂S₃ and 2.00 g of H₂O are reacted? A few of the molar masses are as follows: Al₂S₃= 150.17 g/mol,H₂O = 18.02 g/mol. Al₂S₃(s)+ 6 H₂O(l)→ 2 Al(OH)₃(s)+ 3 H₂S(g)

Accepted Answer

The answer of According to the following reaction,what amount of...

Question 18

Determine the theoretical yield of H₂S (in moles)if 4.0 mol Al₂S₃ and 4.0 mol H₂O are reacted according to the following balanced reaction.A possibly useful molar mass is Al₂S₃= 150.17 g/mol. Al₂S₃(s)+ 6 H₂O(l)→ 2 Al(OH)₃(s)+ 3 H₂S(g)

Accepted Answer

The limiting reactant is H₂O, as 6 moles of H₂O are needed per mole of Al₂S₃. With 4 moles of H₂O, only 2/3 of a mole of Al₂S₃ can react, producing 2 moles of H₂S.

Question 19

Give the theoretical yield,in moles,of CO₂from the reaction of 4.00 moles of C₈H₁₈ with 4.00 moles of O₂. 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

Accepted Answer

According to stoichiometry of the reaction, 2 moles of C₈H₁₈ reacts with 25 moles of O₂ to give 16 moles of CO₂. Thus, for 4.00 moles of C₈H₁₈, the moles of CO₂ produced will be (16/2)*4.00= 2.56 moles. Therefore, the theoretical yield of CO₂ is 2.56 moles. The correct option is (C).

Question 20

Give the theoretical yield,in grams,of CO₂from the reaction of 4.000 moles of C₈H₁₈ with 4.000 moles of O₂. 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

Accepted Answer

The reaction is limited by O₂, as 25 moles of O₂ are needed for 2 moles of C₈H₁₈. With 4 moles of O₂, the reaction produces (4/25) * 16 moles of CO₂, which is 2.56 moles. The molar mass of CO₂ is 44.01 g/mol, so the theoretical yield is 2.56 * 44.01 = 112.7 g.

Quiz 4: Chemical Quantities and Aqueous Reactions

Consider the following balanced reaction.What mass (in g) of CO₂ can be formed from 288 mg of O₂? Assume that there is excess C₃H₇SH present. C₃H₇SH(l) + 6 O₂(g) → 3 CO₂(g) + SO₂(g) + 4 H₂O(g)

How many molecules of H₂S are required to form 79.0 g of sulfur according to the following reaction? Assume excess SO₂. 2 H₂S(g) + SO₂(g) → 3 S(s) + 2H₂O(l)

Determine the theoretical yield of HCl if 60.0 g of BCl₃and 37.5 g of H₂O are reacted according to the following balanced reaction.A possibly useful molar mass is BCl₃ = 117.16 g/mol. BCl₃(g) + 3 H₂O(l) → H₃BO₃(s) + 3 HCl(g)

Consider the following reaction.How many moles of oxygen are required to produce 2.33 moles of water? Assume that there is excess C₃H₇SH present. C₃H₇SH(l) + 6 O₂(g) → 3 CO₂(g) + SO₂(g) + 4 H₂O(g)

A 12.39 g sample of phosphorus reacts with 42.54 g of chlorine to form only phosphorus trichloride (PCl₃) .If it is the only product,what mass of PCl₃ is formed?

Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N₂O₄ and 45.0 g N₂H₄.Some possibly useful molar masses are as follows: N₂O₄ = 92.02 g/mol,N₂H₄ = 32.05 g/mol. N₂O₄(l) + 2 N₂H₄(l) → 3 N₂(g) + 4 H₂O(g)

How many grams of Li₃N can be formed from 1.75 moles of Li? Assume an excess of nitrogen. 6 Li(s) + N₂(g) → 2 Li₃N(s)

How many molecules of HCl are formed when 50.0 g of water reacts according to the following balanced reaction? Assume excess ICl₃. 2 ICl₃ + 3 H₂O → ICl + HIO₃ + 5 HCl

Consider the following balanced reaction.How many grams of water are required to form 75.9 g of HNO₃? Assume that there is excess NO₂ present.The molar masses are as follows: H₂O = 18.02 g/mol,HNO₃ = 63.02 g/mol. 3 NO₂(g) + H₂O(l) → 2 HNO₃(aq) + NO(g)

How many grams of oxygen are formed when 6.21 moles of KOH are formed? 4 KO(s) + 2 H₂O(l) → 4 KOH(s) + O₂(g)

How many moles of oxygen are formed when 58.6 g of KNO₃ decomposes according to the following reaction? The molar mass of KNO₃ is 101.11 g/mol. 4 KNO₃(s) → 2 K₂O(s) + 2 N₂(g) + 5 O₂(g)

According to the following balanced reaction,how many moles of NO are formed from 8.44 moles of NO₂if there is plenty of water present? 3 NO₂(g) + H₂O(l) → 2 HNO₃(aq) + NO(g)

According to the following reaction,how many grams of sulfur are formed when 37.4 g of water are formed? 2 H₂S(g) + SO₂(g) → 3 S(s) + 2H₂O(l)

A 14.01 g sample of N₂ reacts with 3.02 g of H₂ to form ammonia (NH₃) .If ammonia is the only product,what mass of ammonia is formed?

According to the following balanced reaction,how many moles of KO are required to exactly react with 5.44 moles of H₂O? 4 KO(s) + 2 H₂O(l) → 4 KOH(s) + O₂(g)

Carbonic acid can form water and carbon dioxide upon heating.How much carbon dioxide is formed from 6.20 g of carbonic acid? H₂CO₃ → H₂O + CO₂

According to the following reaction,what amount of Al₂S₃ remains when 20.00 g of Al₂S₃ and 2.00 g of H₂O are reacted? A few of the molar masses are as follows: Al₂S₃= 150.17 g/mol,H₂O = 18.02 g/mol. Al₂S₃(s) + 6 H₂O(l) → 2 Al(OH) ₃(s) + 3 H₂S(g)

Determine the theoretical yield of H₂S (in moles) if 4.0 mol Al₂S₃ and 4.0 mol H₂O are reacted according to the following balanced reaction.A possibly useful molar mass is Al₂S₃= 150.17 g/mol. Al₂S₃(s) + 6 H₂O(l) → 2 Al(OH) ₃(s) + 3 H₂S(g)

Give the theoretical yield,in moles,of CO₂from the reaction of 4.00 moles of C₈H₁₈ with 4.00 moles of O₂. 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

Give the theoretical yield,in grams,of CO₂from the reaction of 4.000 moles of C₈H₁₈ with 4.000 moles of O₂. 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

Quiz 4: Chemical Quantities and Aqueous Reactions

Consider the following balanced reaction.What mass (in g) of CO2 can be formed from 288 mg of O2? Assume that there is excess C3H7SH present. C3H7SH(l) + 6 O2(g) → 3 CO2(g) + SO2(g) + 4 H2O(g)

How many molecules of H2S are required to form 79.0 g of sulfur according to the following reaction? Assume excess SO2. 2 H2S(g) + SO2(g) → 3 S(s) + 2H2O(l)

Determine the theoretical yield of HCl if 60.0 g of BCl3 and 37.5 g of H2O are reacted according to the following balanced reaction.A possibly useful molar mass is BCl3 = 117.16 g/mol. BCl3(g) + 3 H2O(l) → H3BO3(s) + 3 HCl(g)

Consider the following reaction.How many moles of oxygen are required to produce 2.33 moles of water? Assume that there is excess C3H7SH present. C3H7SH(l) + 6 O2(g) → 3 CO2(g) + SO2(g) + 4 H2O(g)

A 12.39 g sample of phosphorus reacts with 42.54 g of chlorine to form only phosphorus trichloride (PCl3) .If it is the only product,what mass of PCl3 is formed?

Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N2O4 and 45.0 g N2H4.Some possibly useful molar masses are as follows: N2O4 = 92.02 g/mol,N2H4 = 32.05 g/mol. N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g)

How many grams of Li3N can be formed from 1.75 moles of Li? Assume an excess of nitrogen. 6 Li(s) + N2(g) → 2 Li3N(s)

How many molecules of HCl are formed when 50.0 g of water reacts according to the following balanced reaction? Assume excess ICl3. 2 ICl3 + 3 H2O → ICl + HIO3 + 5 HCl

Consider the following balanced reaction.How many grams of water are required to form 75.9 g of HNO3? Assume that there is excess NO2 present.The molar masses are as follows: H2O = 18.02 g/mol,HNO3 = 63.02 g/mol. 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)

How many grams of oxygen are formed when 6.21 moles of KOH are formed? 4 KO(s) + 2 H2O(l) → 4 KOH(s) + O2(g)

How many moles of oxygen are formed when 58.6 g of KNO3 decomposes according to the following reaction? The molar mass of KNO3 is 101.11 g/mol. 4 KNO3(s) → 2 K2O(s) + 2 N2(g) + 5 O2(g)

According to the following balanced reaction,how many moles of NO are formed from 8.44 moles of NO2 if there is plenty of water present? 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)

According to the following reaction,how many grams of sulfur are formed when 37.4 g of water are formed? 2 H2S(g) + SO2(g) → 3 S(s) + 2H2O(l)

A 14.01 g sample of N2 reacts with 3.02 g of H2 to form ammonia (NH3) .If ammonia is the only product,what mass of ammonia is formed?

According to the following balanced reaction,how many moles of KO are required to exactly react with 5.44 moles of H2O? 4 KO(s) + 2 H2O(l) → 4 KOH(s) + O2(g)

Carbonic acid can form water and carbon dioxide upon heating.How much carbon dioxide is formed from 6.20 g of carbonic acid? H2CO3 → H2O + CO2

According to the following reaction,what amount of Al2S3 remains when 20.00 g of Al2S3 and 2.00 g of H2O are reacted? A few of the molar masses are as follows: Al2S3 = 150.17 g/mol,H2O = 18.02 g/mol. Al2S3(s) + 6 H2O(l) → 2 Al(OH) 3(s) + 3 H2S(g)

Determine the theoretical yield of H2S (in moles) if 4.0 mol Al2S3 and 4.0 mol H2O are reacted according to the following balanced reaction.A possibly useful molar mass is Al2S3 = 150.17 g/mol. Al2S3(s) + 6 H2O(l) → 2 Al(OH) 3(s) + 3 H2S(g)

Give the theoretical yield,in moles,of CO2 from the reaction of 4.00 moles of C8H18 with 4.00 moles of O2. 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

Give the theoretical yield,in grams,of CO2 from the reaction of 4.000 moles of C8H18 with 4.000 moles of O2. 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

Consider the following balanced reaction.What mass (in g) of CO₂ can be formed from 288 mg of O₂? Assume that there is excess C₃H₇SH present. C₃H₇SH(l) + 6 O₂(g) → 3 CO₂(g) + SO₂(g) + 4 H₂O(g)

How many molecules of H₂S are required to form 79.0 g of sulfur according to the following reaction? Assume excess SO₂. 2 H₂S(g) + SO₂(g) → 3 S(s) + 2H₂O(l)

Determine the theoretical yield of HCl if 60.0 g of BCl₃and 37.5 g of H₂O are reacted according to the following balanced reaction.A possibly useful molar mass is BCl₃ = 117.16 g/mol. BCl₃(g) + 3 H₂O(l) → H₃BO₃(s) + 3 HCl(g)

Consider the following reaction.How many moles of oxygen are required to produce 2.33 moles of water? Assume that there is excess C₃H₇SH present. C₃H₇SH(l) + 6 O₂(g) → 3 CO₂(g) + SO₂(g) + 4 H₂O(g)

A 12.39 g sample of phosphorus reacts with 42.54 g of chlorine to form only phosphorus trichloride (PCl₃) .If it is the only product,what mass of PCl₃ is formed?

Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N₂O₄ and 45.0 g N₂H₄.Some possibly useful molar masses are as follows: N₂O₄ = 92.02 g/mol,N₂H₄ = 32.05 g/mol. N₂O₄(l) + 2 N₂H₄(l) → 3 N₂(g) + 4 H₂O(g)

How many grams of Li₃N can be formed from 1.75 moles of Li? Assume an excess of nitrogen. 6 Li(s) + N₂(g) → 2 Li₃N(s)

How many molecules of HCl are formed when 50.0 g of water reacts according to the following balanced reaction? Assume excess ICl₃. 2 ICl₃ + 3 H₂O → ICl + HIO₃ + 5 HCl

Consider the following balanced reaction.How many grams of water are required to form 75.9 g of HNO₃? Assume that there is excess NO₂ present.The molar masses are as follows: H₂O = 18.02 g/mol,HNO₃ = 63.02 g/mol. 3 NO₂(g) + H₂O(l) → 2 HNO₃(aq) + NO(g)

How many grams of oxygen are formed when 6.21 moles of KOH are formed? 4 KO(s) + 2 H₂O(l) → 4 KOH(s) + O₂(g)

How many moles of oxygen are formed when 58.6 g of KNO₃ decomposes according to the following reaction? The molar mass of KNO₃ is 101.11 g/mol. 4 KNO₃(s) → 2 K₂O(s) + 2 N₂(g) + 5 O₂(g)

According to the following balanced reaction,how many moles of NO are formed from 8.44 moles of NO₂if there is plenty of water present? 3 NO₂(g) + H₂O(l) → 2 HNO₃(aq) + NO(g)

According to the following reaction,how many grams of sulfur are formed when 37.4 g of water are formed? 2 H₂S(g) + SO₂(g) → 3 S(s) + 2H₂O(l)

A 14.01 g sample of N₂ reacts with 3.02 g of H₂ to form ammonia (NH₃) .If ammonia is the only product,what mass of ammonia is formed?

According to the following balanced reaction,how many moles of KO are required to exactly react with 5.44 moles of H₂O? 4 KO(s) + 2 H₂O(l) → 4 KOH(s) + O₂(g)

Carbonic acid can form water and carbon dioxide upon heating.How much carbon dioxide is formed from 6.20 g of carbonic acid? H₂CO₃ → H₂O + CO₂

According to the following reaction,what amount of Al₂S₃ remains when 20.00 g of Al₂S₃ and 2.00 g of H₂O are reacted? A few of the molar masses are as follows: Al₂S₃= 150.17 g/mol,H₂O = 18.02 g/mol. Al₂S₃(s) + 6 H₂O(l) → 2 Al(OH) ₃(s) + 3 H₂S(g)

Determine the theoretical yield of H₂S (in moles) if 4.0 mol Al₂S₃ and 4.0 mol H₂O are reacted according to the following balanced reaction.A possibly useful molar mass is Al₂S₃= 150.17 g/mol. Al₂S₃(s) + 6 H₂O(l) → 2 Al(OH) ₃(s) + 3 H₂S(g)

Give the theoretical yield,in moles,of CO₂from the reaction of 4.00 moles of C₈H₁₈ with 4.00 moles of O₂. 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

Give the theoretical yield,in grams,of CO₂from the reaction of 4.000 moles of C₈H₁₈ with 4.000 moles of O₂. 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O