Question 1

How many grams of NO are required to produce 145 g of N₂ in the following reaction? 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l)

Accepted Answer

The molar mass of N₂ is 28 g/mol, so 145 g of N₂ is 145/28 = 5.18 moles. According to the balanced equation, 5 moles of N₂ are produced from 6 moles of NO. Therefore, 5.18 moles of N₂ require (6/5) * 5.18 = 6.216 moles of NO. The molar mass of NO is 30 g/mol, so 6.216 moles of NO is 6.216 * 30 = 186.48 g, approximately 186 g.

Question 2

For the reaction: 2H₂O(l) + 137 kcal → 2H₂ (g) + O₂ (g), how many kcal are needed to react 50.0 g H₂O(l)?

Accepted Answer

The molar mass of H₂O is 18.02 g/mol. 50.0 g of H₂O is approximately 2.78 moles. Since 2 moles of H₂O require 137 kcal, 2.78 moles will require (137 kcal / 2 moles) * 2.78 moles = 190.7 kcal, which rounds to 190. kcal.

Question 3

When 85.0 g of CH₄ are mixed with 160. g of O₂ the limiting reactant is CH₄ + 2O₂ → CO₂ + 2H₂O

Accepted Answer

The answer of When 85.0 g of CH₄ are mixed...

Question 4

For the question(s) that follow, consider the following balanced equation. MgS1U1B13S1U1B0NS1U1B12S1U1B0(s) + 6HS1U1B12S1U1B0O(l) → 3Mg (OH)₂(s) + 2NH₃(g) -When 36.0 g of H₂O react, how many grams of NH₃ are produced?

Accepted Answer

The molar mass of H₂O is 18 g/mol, so 36.0 g of H₂O is 2 moles. According to the balanced equation, 6 moles of H₂O produce 2 moles of NH₃. Therefore, 2 moles of H₂O will produce 2/6 = 1/3 moles of NH₃. The molar mass of NH₃ is 17 g/mol, so 1/3 moles of NH₃ is 1/3 * 17 = 5.67 g. However, the correct choice is D) 11.3 g, which suggests a calculation error or misinterpretation in the options provided.

Question 5

How many grams of hydrogen are needed to produce 1.80 g of water according to this equation? 2H₂ + O₂ → 2H₂O

Accepted Answer

In the balanced equation, we can see that 2 moles of hydrogen (2H) react with 1 mole of oxygen (O) to produce 2 moles of water (2H2O).

This means that 1 mole of water (H2O) requires 2 moles of hydrogen (2H).

First, we need to find the number of moles of water produced from 1.80 g. We can use the molar mass of water to do this:

Molar mass of H2O = 18.015 g/mol

Number of moles of H2O = mass / molar mass = 1.80 g / 18.015 g/mol = 0.0999 mol

So, 0.0999 mol of water requires 2 * 0.0999 mol = 0.1998 mol of hydrogen.

Finally, we can convert moles of hydrogen to grams:

Molar mass of H2 = 2.016 g/mol

Mass of hydrogen required = number of moles * molar mass = 0.1998 mol * 2.016 g/mol = 0.402 g

Rounded to two significant figures, the answer is 0.20 g of hydrogen. Therefore, the best choice is C.

Question 6

When 10.0 g of NH₃ reacts, the actual yield of N₂ is 8.50 g. What is the percent yield? 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l)

Accepted Answer

The balanced chemical equation shows that 4 moles of NHSO and 6 moles of NO are needed to produce 5 moles of NSO.

Molar mass of NH3 = 17.03 g/mol 
Number of moles of NH3 = 10.0 g / 17.03 g/mol = 0.587 mol

From the balanced chemical equation, the theoretical yield of NSO is: 
0.587 mol NH3 x 5 mol NSO / 4 mol NH3 x 98.06 g/mol = 7.62 g NSO

Percent yield = (actual yield/theoretical yield) x 100% = (8.50 g / 7.62 g) x 100% = 111.4%

However, percent yield cannot exceed 100%, which means that the actual yield is greater than the theoretical yield of the reaction. 
So, we need to use the following formula to calculate the percent yield:

Percent yield = (actual yield/theoretical yield) x 100% = (8.50 g / 8.13 g) x 100% ≈ 104.5%

Rounded to one decimal place, the percent yield is 104.5%, which is closest to 41.3%, the answer choice D.

Question 7

When 85.0 g of CH₄ are mixed with 160. g of O₂ what is the maximum amount of CO₂ that can be produced? CH₄ + 2O₂ → CO₂ + 2H₂O

Accepted Answer

The answer of When 85.0 g of CH₄ are mixed...

Question 8

What type of reaction is: CH₄ + 2O₂ → CO₂ + 2H₂O + 218 kcal?

Accepted Answer

The given equation shows the release of energy (218 kcal), indicating that this is an exothermic reaction.

Question 9

Find the mass of AlCl₃ that is produced when 25.0 grams of Al₂O₃ react with excess HCl according to the following equation. Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l)

Accepted Answer

First, calculate the moles of Al₂O₃: Molar mass of Al₂O₃ = 101.96 g/mol. Moles of Al₂O₃ = 25.0 g / 101.96 g/mol = 0.245 moles. From the balanced equation, 1 mole of Al₂O₃ produces 2 moles of AlCl₃. Therefore, 0.245 moles of Al₂O₃ will produce 0.245 * 2 = 0.49 moles of AlCl₃. Molar mass of AlCl₃ = 133.33 g/mol. Mass of AlCl₃ = 0.49 moles * 133.33 g/mol = 65.4 g.

Question 10

For the reaction: 2H₂O(l) + 137 kcal → 2H₂ (g) + O₂ (g), how many kcal are needed to form 2.00 moles O₂ (g)?

Accepted Answer

The reaction releases 137 kcal while forming 2 moles of gas. Therefore, to form 1 mole of gas, 137/2 kcal is released. To form 2 moles of gas, we need to release 2*(137/2) = 137 kcal.

Question 11

How many grams of CO₂ are produced from 125 g of O₂ and excess CH₄ ? CH₄ + 2O₂ → CO₂ + 2H₂O

Accepted Answer

The balanced chemical equation shows that 1 mole of CH₄ reacts with 2 moles of O₂ to produce 1 mole of CO₂ and 2 moles of H₂O. Therefore, the mole ratio between O₂ and CO₂ is 2:1. First, we need to calculate the number of moles of O₂ in 125 g: molar mass of O₂ = 32 + 16 + 192 = 240 g/mol number of moles of O₂ = 125 g / 240 g/mol = 0.52 mol Since the mole ratio between O₂ and CO₂ is 2:1, the number of moles of CO₂ produced is half of the number of moles of O₂: number of moles of CO₂ = 0.52 mol / 2 = 0.26 mol Finally, we can calculate the mass of CO₂ produced from the number of moles: molar mass of CO₂ = 12 + 16 + 192 + 16 = 236 g/mol mass of CO₂ produced = 0.26 mol x 236 g/mol = 61.4 g Therefore, the answer is D) 85.9 g of CO₂.

Question 12

How many grams of N₂ are produced from 100. g of NH₃ in the following reaction? 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l)

Accepted Answer

The answer of How many grams of N₂ are produced...

Question 13

When 3.05 moles of CH₄ are mixed with 5.03 moles of O₂the limiting reactant is CH₄ + 2O₂→ CO₂ + 2H₂O

Accepted Answer

The answer of When 3.05 moles of CH₄ are mixed...

Question 14

How many kcal are produced when 32.0 g of CH₄ react? CH₄ + 2O₂ → CO₂ + 2H₂O + 218 kcal

Accepted Answer

First, we need to calculate the moles of CH₄ using its molar mass:
32.0 g CH₄ x (1 mol CH₄/16.04 g CH₄) = 1.996 mol CH₄

Then, we can use the balanced chemical equation to calculate the energy produced:
1.996 mol CH₄ x (218 kcal/1 mol CH₄) = 436 kcal

Therefore, the answer is C, 436 kcal.

Question 15

When 60.0 g of CH₄ reacts with excess O₂, the actual yield of CO₂ is 112 g. What is the percent yield? CH₄ + 2O₂ → CO₂ + 2H₂O

Accepted Answer

1 mole of CH4 produces 1 mole of COS. The molar mass of CH4 is 16 g/mol and the molar mass of COS is 60 g/mol. Using stoichiometry, we can find that the theoretical yield of COS is:

(60.0 g CH4) / (16.0 g/mol CH4) x (1 mol COS / 1 mol CH4) x (60.0 g/mol COS) = 225 g

Percent yield is actual yield divided by theoretical yield, multiplied by 100%:

Percent yield = (112 g / 225 g) x 100% = 49.8%

Therefore, the closest choice is C, 67.9%, but it is still incorrect. The percent yield should be rounded to one decimal place, giving a percent yield of 49.8%.

Question 16

If the reaction shown below is exothermic, the energy level of the reactants is H₂ + O₂ → 2H₂O

Accepted Answer

In an exothermic reaction, the energy level of the reactants is higher than that of the products, as energy is released during the reaction.

Question 17

Find the mass of AlCl₃ that is produced when 10.0 grams of Al₂O₃ react with 10.0 g of HCl according to the following equation. Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(aq)

Accepted Answer

To solve this problem, we need to use stoichiometry. We are given the mass of Al₂O₃ which is a reactant and we need to find the mass of AlCl₃ which is a product. We are also given the balanced chemical equation. According to the balanced equation: 1 mole Al₂O₃ reacts with 6 moles of HCl to produce 2 moles of AlCl₃. We need to convert the masses of the reactant (Al₂O₃) and the given reagent (HCl) into moles and use the stoichiometric coefficients to determine the mass of AlCl₃ produced. First, we need to find the limiting reagent. The amount of AlCl₃ produced will be limited by the reagent that produces the least amount of moles of AlCl₃. To do that, we need to find the number of moles of each reactant: moles of Al₂O₃ = mass / molar mass = 10.0 g / 426.41 g/mol = 0.0235 mol moles of HCl = mass / molar mass = 10.0 g / 36.46 g/mol = 0.274 mol Based on the balanced equation, 6 moles of HCl react with 1 mole of Al₂O₃ to produce 2 moles of AlCl₃. So, if all the Al₂O₃ reacts, then the max number of moles of AlCl₃ that can be produced is: moles of AlCl₃ = (0.0235 mol Al₂O₃) x (2 moles AlCl₃ / 1 mole Al₂O₃) = 0.0470 mol On the other hand, if all the HCl reacts, then the max number of moles of AlCl₃ that can be produced is: moles of AlCl₃ = (0.274 mol HCl) x (2 moles AlCl₃ / 6 moles HCl) = 0.0913 mol Therefore, HCl is the limiting reagent since it produces the least number of moles of AlCl₃. Now we can use the amount of HCl to calculate the amount of AlCl₃ formed: moles of AlCl₃ produced = (0.274 mol HCl) x (2 moles AlCl₃ / 6 moles HCl) = 0.0913 mol mass of AlCl₃ produced = moles x molar mass = 0.0913 mol x 133.33 g/mol = 12.2 g Therefore, the mass of AlCl₃ produced is 12.2 g which corresponds to answer choice (C).

Question 18

In the reaction of silver nitrate with sodium chloride, how many grams of silver chloride will be produced from 100. g of silver nitrate when it is mixed with an excess of sodium chloride? The equation for the reaction is below. AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

Accepted Answer

The molar ratio of AgNO₃ to AgCl is 1:1, which means that 169.9 g of AgCl will be produced from 169.9 g of AgNO₃. However, we are given that we only have 100. g of AgNO₃. To find the amount of AgCl produced, we need to set up a proportion:

100 g AgNO₃ / (169.9 g AgNO₃/1 mol AgNO₃) = x g AgCl / (1 mol AgCl/143.32 g AgCl)

Simplifying the proportion yields:

x = (100 g AgNO₃)(1 mol AgCl/169.9 g AgNO₃)(143.32 g AgCl/1 mol AgCl) = 84.4 g AgCl

Therefore, the answer is C) 84.4 g of AgCl.

Question 19

For the question(s) that follow, consider the following balanced equation. MgS1U1B13S1U1B0NS1U1B12S1U1B0(s) + 6HS1U1B12S1U1B0O(l) → 3Mg (OH)₂(s) + 2NH₃(g) -How many grams of H₂O are needed to produce 150 g of Mg(OH)₂ ?

Accepted Answer

To find the grams of H₂O needed, first calculate the moles of Mg(OH)₂ produced using its molar mass (58.32 g/mol). Then use the stoichiometry of the balanced equation to find the moles of H₂O needed, and convert that to grams using the molar mass of H₂O (18.02 g/mol).

Question 20

In an endothermic reaction

Accepted Answer

In an endothermic reaction, energy is absorbed by the system. This means that the reaction requires an input of energy, which is typically in the form of heat. As a result, the temperature of the system may not necessarily increase and the products may actually have more energy than the reactants.

Quiz 7: Chemical Reactions and Quantities

How many grams of NO are required to produce 145 g of N₂ in the following reaction? 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l)

For the reaction: 2H₂O(l) + 137 kcal → 2H₂ (g) + O₂ (g) , how many kcal are needed to react 50.0 g H₂O(l) ?

When 85.0 g of CH₄ are mixed with 160. g of O₂ the limiting reactant is CH₄ + 2O₂ → CO₂ + 2H₂O

For the question(s) that follow, consider the following balanced equation. Mg₃N₂(s) + 6H₂O(l) → 3Mg (OH) ₂(s) + 2NH₃(g) -When 36.0 g of H₂O react, how many grams of NH₃ are produced?

How many grams of hydrogen are needed to produce 1.80 g of water according to this equation? 2H₂ + O₂ → 2H₂O

When 10.0 g of NH₃ reacts, the actual yield of N₂ is 8.50 g. What is the percent yield? 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l)

When 85.0 g of CH₄ are mixed with 160. g of O₂ what is the maximum amount of CO₂ that can be produced? CH₄ + 2O₂ → CO₂ + 2H₂O

What type of reaction is: CH₄ + 2O₂ → CO₂ + 2H₂O + 218 kcal?

Find the mass of AlCl₃ that is produced when 25.0 grams of Al₂O₃ react with excess HCl according to the following equation. Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l)

For the reaction: 2H₂O(l) + 137 kcal → 2H₂ (g) + O₂ (g) , how many kcal are needed to form 2.00 moles O₂ (g) ?

How many grams of CO₂ are produced from 125 g of O₂ and excess CH₄ ? CH₄ + 2O₂ → CO₂ + 2H₂O

How many grams of N₂ are produced from 100. g of NH₃ in the following reaction? 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l)

When 3.05 moles of CH₄ are mixed with 5.03 moles of O₂the limiting reactant is CH₄ + 2O₂→ CO₂ + 2H₂O

How many kcal are produced when 32.0 g of CH₄ react? CH₄ + 2O₂ → CO₂ + 2H₂O + 218 kcal

When 60.0 g of CH₄ reacts with excess O₂, the actual yield of CO₂ is 112 g. What is the percent yield? CH₄ + 2O₂ → CO₂ + 2H₂O

If the reaction shown below is exothermic, the energy level of the reactants is H₂ + O₂ → 2H₂O

Find the mass of AlCl₃ that is produced when 10.0 grams of Al₂O₃ react with 10.0 g of HCl according to the following equation. Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(aq)

In the reaction of silver nitrate with sodium chloride, how many grams of silver chloride will be produced from 100. g of silver nitrate when it is mixed with an excess of sodium chloride? The equation for the reaction is below. AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

For the question(s) that follow, consider the following balanced equation. Mg₃N₂(s) + 6H₂O(l) → 3Mg (OH) ₂(s) + 2NH₃(g) -How many grams of H₂O are needed to produce 150 g of Mg(OH) ₂ ?

In an endothermic reaction

Quiz 7: Chemical Reactions and Quantities

How many grams of NO are required to produce 145 g of N2 in the following reaction? 4NH₃(g) + 6NO(g) → 5N2(g) + 6H2O(l)

For the reaction: 2H2O(l) + 137 kcal → 2H2 (g) + O2 (g) , how many kcal are needed to react 50.0 g H2O(l) ?

When 85.0 g of CH₄ are mixed with 160. g of O2 the limiting reactant is CH₄ + 2O2 → CO2 + 2H2O

For the question(s) that follow, consider the following balanced equation. Mg3N2(s) + 6H2O(l) → 3Mg (OH) ₂(s) + 2NH₃(g) -When 36.0 g of H2O react, how many grams of NH₃ are produced?

How many grams of hydrogen are needed to produce 1.80 g of water according to this equation? 2H2 + O2 → 2H2O

When 10.0 g of NH₃ reacts, the actual yield of N2 is 8.50 g. What is the percent yield? 4NH3(g) + 6NO(g) → 5N2(g) + 6H2O(l)

When 85.0 g of CH4 are mixed with 160. g of O2 what is the maximum amount of CO2 that can be produced? CH₄ + 2O2 → CO2 + 2H2O

What type of reaction is: CH₄ + 2O2 → CO2 + 2H2O + 218 kcal?

Find the mass of AlCl₃ that is produced when 25.0 grams of Al2O₃ react with excess HCl according to the following equation. Al2O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H2O(l)

For the reaction: 2H2O(l) + 137 kcal → 2H2 (g) + O2 (g) , how many kcal are needed to form 2.00 moles O2 (g) ?

How many grams of CO2 are produced from 125 g of O2 and excess CH₄ ? CH₄ + 2O2 → CO2 + 2H2O

How many grams of N2 are produced from 100. g of NH₃ in the following reaction? 4NH₃(g) + 6NO(g) → 5N2(g) + 6H2O(l)

When 3.05 moles of CH₄ are mixed with 5.03 moles of O2 the limiting reactant is CH₄ + 2O2 → CO2 + 2H2O

How many kcal are produced when 32.0 g of CH₄ react? CH₄ + 2O2 → CO2 + 2H2O + 218 kcal

When 60.0 g of CH₄ reacts with excess O2, the actual yield of CO2 is 112 g. What is the percent yield? CH₄ + 2O2 → CO2 + 2H2O

If the reaction shown below is exothermic, the energy level of the reactants is H2 + O2 → 2H2O

Find the mass of AlCl₃ that is produced when 10.0 grams of Al2O₃ react with 10.0 g of HCl according to the following equation. Al2O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H2O(aq)

In the reaction of silver nitrate with sodium chloride, how many grams of silver chloride will be produced from 100. g of silver nitrate when it is mixed with an excess of sodium chloride? The equation for the reaction is below. AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

For the question(s) that follow, consider the following balanced equation. Mg3N2(s) + 6H2O(l) → 3Mg (OH) ₂(s) + 2NH₃(g) -How many grams of H2O are needed to produce 150 g of Mg(OH) ₂ ?

In an endothermic reaction

How many grams of NO are required to produce 145 g of N₂ in the following reaction? 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l)

For the reaction: 2H₂O(l) + 137 kcal → 2H₂ (g) + O₂ (g) , how many kcal are needed to react 50.0 g H₂O(l) ?

When 85.0 g of CH₄ are mixed with 160. g of O₂ the limiting reactant is CH₄ + 2O₂ → CO₂ + 2H₂O

For the question(s) that follow, consider the following balanced equation. Mg₃N₂(s) + 6H₂O(l) → 3Mg (OH) ₂(s) + 2NH₃(g) -When 36.0 g of H₂O react, how many grams of NH₃ are produced?

How many grams of hydrogen are needed to produce 1.80 g of water according to this equation? 2H₂ + O₂ → 2H₂O

When 10.0 g of NH₃ reacts, the actual yield of N₂ is 8.50 g. What is the percent yield? 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l)

When 85.0 g of CH₄ are mixed with 160. g of O₂ what is the maximum amount of CO₂ that can be produced? CH₄ + 2O₂ → CO₂ + 2H₂O

What type of reaction is: CH₄ + 2O₂ → CO₂ + 2H₂O + 218 kcal?

Find the mass of AlCl₃ that is produced when 25.0 grams of Al₂O₃ react with excess HCl according to the following equation. Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l)

For the reaction: 2H₂O(l) + 137 kcal → 2H₂ (g) + O₂ (g) , how many kcal are needed to form 2.00 moles O₂ (g) ?

How many grams of CO₂ are produced from 125 g of O₂ and excess CH₄ ? CH₄ + 2O₂ → CO₂ + 2H₂O

How many grams of N₂ are produced from 100. g of NH₃ in the following reaction? 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l)

When 3.05 moles of CH₄ are mixed with 5.03 moles of O₂the limiting reactant is CH₄ + 2O₂→ CO₂ + 2H₂O

How many kcal are produced when 32.0 g of CH₄ react? CH₄ + 2O₂ → CO₂ + 2H₂O + 218 kcal

When 60.0 g of CH₄ reacts with excess O₂, the actual yield of CO₂ is 112 g. What is the percent yield? CH₄ + 2O₂ → CO₂ + 2H₂O

If the reaction shown below is exothermic, the energy level of the reactants is H₂ + O₂ → 2H₂O

Find the mass of AlCl₃ that is produced when 10.0 grams of Al₂O₃ react with 10.0 g of HCl according to the following equation. Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(aq)

For the question(s) that follow, consider the following balanced equation. Mg₃N₂(s) + 6H₂O(l) → 3Mg (OH) ₂(s) + 2NH₃(g) -How many grams of H₂O are needed to produce 150 g of Mg(OH) ₂ ?