Question 1

Consider an initial mixture of CH₄ and O₂ represented in the container below: Given the reaction CH₄ + 2O₂ → CO₂ + 2H₂O,which of the following represents a stoichiometric picture of the container after the reaction has gone to completion? A) B) C) D) E)none of the above

Accepted Answer

The reaction shows that for every one molecule of CH₄, two molecules of O₂ are required. After the reaction completes, all of the CH₄ will have been used up, and the amount of O₂ will have decreased by a factor of two while the amount of the products, CO₂ and 2H₂O, will have increased by the same factor. Thus, the stoichiometric picture should show no more CH₄, half as much O₂, and twice as much of each product. This matches choice C.

Question 2

A compound contains 43.84 % carbon atoms,3.65% hydrogen atoms,and 8.68 % fluorine atoms by mass.Each molecule of this compound contains one fluorine atom.What is the total number of carbon,hydrogen,and fluorine atoms in one molecule of this compound?
A)17
B)11
C)10
D)7
E)14

Accepted Answer

The compound contains 43.84% carbon, 3.65% hydrogen, and 8.68% fluorine by mass. Given that there is one fluorine atom per molecule, we can use the atomic masses (approximately C = 12, H = 1, F = 19) to find the ratio of atoms in the molecule. The mass percentages suggest the molecule is primarily composed of carbon and hydrogen, with a single fluorine atom. The percentages don't directly translate to the number of atoms because they are based on mass, not count. However, knowing there's one fluorine atom, we can infer the relative amounts of carbon and hydrogen needed to reach the given mass percentages. Given the high percentage of carbon, a rough estimation would suggest multiple carbon atoms, likely in a higher number than hydrogen atoms due to carbon's larger atomic mass. The only option that reasonably fits the criteria of having multiple carbons, more than one hydrogen, and exactly one fluorine, and sums up to a total that could reflect the given mass percentages when considering their atomic masses, is 17. This suggests a molecule with a composition that could include, for example, around 8 carbons (8*12 = 96, close to 43.84%), around 8 hydrogens (8*1 = 8, close to 3.65%), and 1 fluorine (1*19 = 19, close to 8.68%), though the exact composition in terms of atom count might vary, the total count aligns with the option given.

Question 3

The products of the combustion of acetone with oxygen are shown in the following equation: __ CH₃COCH₃ + __ O₂ → __ CO₂ + __ H₂O When properly balanced,the equation indicates that ____ mol of O₂ are required for each mole of CH₃COCH₃. A)4 B)6 C)3.5 D)7 E)2

Accepted Answer

The equation is balanced as follows:CH3COCH3 + 4O2 → 3CO2 + 3H2OTherefore, 4 mol of O2 are required for each mole of CH3COCH3.

Question 4

One step in the isolation of pure rhodium metal (Rh)is the precipitation of rhodium(III)hydroxide from a solution containing rhodium(III)sulfate according to the following balanced chemical equation: Rh₂(SO₄)₃(aq)+ 6NaOH(aq)→ 2Rh(OH)₃(s)+ 3Na₂SO₄(aq) What mass of sodium hydroxide is required to precipitate 68.0 g of rhodium(III)hydroxide from a solution containing excess rhodium(III)sulfate? A)5.88 g B)53 g C)17.6 g D)204 g E)68 g

Accepted Answer

According to the balanced chemical equation, 1 mole of Rh(OH)3 is produced for every 6 moles of NaOH used.
We can first convert the given mass of Rh(OH)3 to moles:
68.0 g Rh(OH)3 × (1 mol Rh(OH)3/159.91 g Rh(OH)3) = 0.425 mol Rh(OH)3

Since there is excess Rh2(SO4)3, we can assume that all of the Rh2(SO4)3 is converted to Rh(OH)3, and thus we can use stoichiometry to find the amount of NaOH required.

0.425 mol Rh(OH)3 × (6 mol NaOH/1 mol Rh(OH)3) × (40.00 g NaOH/1 mol NaOH) = 102 g NaOH (to 3 significant figures)

Therefore, the mass of NaOH required to precipitate 68.0 g of Rh(OH)3 is 102 g, which corresponds to choice (B).

Question 5

Ammonia,NH₃,and oxygen can be reacted together in the presence of a catalyst to form only nitrogen monoxide and water.The number of moles of oxygen consumed for every 1.00 mole of NO produced is . A)1.25 B)5 C)3.75 D)0.625 E)2.5

Accepted Answer

The balanced chemical equation for the reaction is 4NH₃ + 5O₂ → 4NO + 6H₂O. For every 4 moles of NO produced, 5 moles of O₂ are consumed. Therefore, for every 1 mole of NO produced, 1.25 moles of O₂ are consumed (5/4 = 1.25).

Question 6

Complete combustion of a 0.40-mol sample of a hydrocarbon,C_xH_y,gives 2.40 mol of CO₂ and 1.60 mol of H₂O.The molecular formula of the original hydrocarbon is A)C₂H₈. B)C₆H₈. C)C₉H₁₆. D)C₅H₅. E)C₄H₇.

Accepted Answer

The answer of Complete combustion of a 0.40-mol sample of...

Question 7

Balance the following expression: __ CH₃CH₂COOH + __ O₂ → __ CO₂ + __ H₂O How many moles of O₂ are required for the complete combustion of 6 mol of propanoic acid? A)6 mol B)33 mol C)21 mol D)38 mol E)4 mol

Accepted Answer

The balanced chemical equation for the combustion of propanoic acid (C3H6O2) is:$$ C_3H_6O_2 + 4.5O_2 \rightarrow 3CO_2 + 3H_2O $$Thus, for every mole of propanoic acid, 4.5 moles of $O_2$ are required. For 6 moles of propanoic acid, $6 \times 4.5 = 27$ moles of $O_2$ are needed.

Question 8

A chemical reaction has the equation: 2A + B → C.In which case is B the limiting reactant? 
A)II
B)I
C)III
D)IV
E)none of these

Accepted Answer

The answer of A chemical reaction has the equation: 2A...

Question 9

The balanced equation for the combustion of ethanol is 2C₂H₅OH(g)+ 7O₂(g)→ 4CO₂(g)+ 6H₂O(g) How many grams of dioxygen are required to burn 3.4 g of C₂H₅OH? A)8.2 g B)11 g C)2.3 g D)38 g E)57 g

Accepted Answer

The balanced equation shows that 2 moles of ethanol react with 7 moles of dioxygen. First, calculate the molar mass of ethanol (C2H5OH): 2(12.01) + 6(1.01) + 16.00 = 46.08 g/mol. Then, find how many moles 3.4 g of ethanol is: 3.4 g / 46.08 g/mol = 0.0738 mol. Since the reaction requires 7 moles of O2 for every 2 moles of ethanol, the moles of O2 needed are (7/2) * 0.0738 = 0.2589 mol. The molar mass of O2 is 32.00 g/mol, so the grams of O2 required are 0.2589 mol * 32.00 g/mol = 8.28 g, which rounds to 8.2 g.

Question 10

One step in the isolation of pure rhodium metal (Rh)is the precipitation of rhodium(III)hydroxide from a solution containing rhodium(III)sulfate according to the following balanced chemical equation: Rh₂(SO₄)₃(aq)+ 6NaOH(aq)→ 2Rh(OH)₃(s)+ 3Na₂SO₄(aq) If 0.730 g of rhodium(III)hydroxide is produced,what mass of sodium sulfate is also produced? A)0.673 g B)1.09 g C)1.01 g D)0.73 g E)0.449 g

Accepted Answer

First, calculate the molar mass of Rh(OH)₃: Rh = 102.91 g/mol, O = 16.00 g/mol, H = 1.01 g/mol. So, Rh(OH)₃ = 102.91 + (16.00*3) + (1.01*3) = 165.94 g/mol. Given 0.730 g of Rh(OH)₃, moles of Rh(OH)₃ = 0.730 g / 165.94 g/mol = 0.0044 mol. From the balanced equation, 2 moles of Rh(OH)₃ produce 3 moles of Na₂SO₄. So, 0.0044 mol of Rh(OH)₃ will produce (0.0044 mol * 3/2) = 0.0066 mol of Na₂SO₄. The molar mass of Na₂SO₄ = 2*22.99 + 32.06 + 4*16.00 = 142.04 g/mol. Mass of Na₂SO₄ = 0.0066 mol * 142.04 g/mol = 0.937 g, which rounds to approximately 1.01 g.

Question 11

One step in the isolation of pure rhodium metal (Rh)is the precipitation of rhodium(III)hydroxide from a solution containing rhodium(III)sulfate according to the following balanced chemical equation: Rh₂(SO₄)₃(aq)+ 6NaOH(aq)→ 2Rh(OH)₃(s)+ 3Na₂SO₄(aq) If 3.20 g of rhodium(III)sulfate reacts with excess sodium hydroxide,what mass of rhodium(III)hydroxide may be produced? A)1.99 g B)6.4 g C)3.2 g D)0.498 g E)3.98 g

Accepted Answer

To solve this, first, calculate the molar mass of rhodium(III) sulfate, $\text{Rh}_2(\text{SO}_4)_3$, and rhodium(III) hydroxide, $\text{Rh(OH)}_3$. The molar mass of $\text{Rh}_2(\text{SO}_4)_3$ is approximately 482.96 g/mol (Rh = 102.91 g/mol, S = 32.07 g/mol, O = 16.00 g/mol), and the molar mass of $\text{Rh(OH)}_3$ is approximately 165.93 g/mol. According to the balanced equation, 2 moles of $\text{Rh(OH)}_3$ are produced per mole of $\text{Rh}_2(\text{SO}_4)_3$. $$3.20 \text{ g} \times \frac{1 \text{ mol}}{482.96 \text{ g}} \times 2 \text{ mol Rh(OH)}_3 \times 165.93 \text{ g/mol} \approx 1.99 \text{ g}$$This calculation shows that 1.99 g of rhodium(III) hydroxide may be produced.

Question 12

2Al(s)+ 6HCl(aq)→ 2AlCl₃(aq)+ 3H₂(g) According to the equation above,how many grams of aluminum are needed to completely react with 4.32 mol of hydrochloric acid? A)350 g B)52.6 g C)38.9 g D)4.32 g E)116.6 g

Accepted Answer

The balanced equation shows that 2 moles of Al react with 6 moles of HCl. Therefore, 1 mole of Al reacts with 3 moles of HCl. For 4.32 moles of HCl, you need 4.32/3 = 1.44 moles of Al. The molar mass of Al is approximately 26.98 g/mol, so 1.44 moles of Al is 1.44 * 26.98 = 38.9 g.

Question 13

Pure copper may be produced by the reaction of copper(I)sulfide with oxygen gas as follows: Cu₂S(s)+ O₂(g)→ 2Cu(s)+ SO₂(g) If 0.190 kg of copper(I)sulfide reacts with excess oxygen,what mass of copper metal may be produced? A)0.19 kg B)0.0379 kg C)0.0758 kg D)0.151 kg E)0.38 kg

Accepted Answer

The balanced equation tells us that 1 mole of CuS reacts with 1 mole of O2 to produce 2 moles of Cu.

First, we need to convert 0.190 kg CuS to moles:
0.190 kg CuS x (1 mol CuS/ 95.61 g CuS) = 0.00199 mol CuS

From the balanced equation, we can see that we will produce twice the amount of moles of copper as moles of CuS reacted. So,

0.00199 mol CuS x (2 mol Cu / 1 mol CuS) = 0.00397 mol Cu

Finally, we can convert moles of Cu to mass:

0.00397 mol Cu x (63.55 g Cu/1 mol Cu) = 0.252 g Cu

Converting to kg, we get 0.151 kg Cu. Therefore, the correct answer is D.

Question 14

A molecular compound contains 92.3 % carbon and 7.7 % hydrogen by mass.If 0.142 mol of the compound weighs 11.08 g,what is its molecular formula? A)C₈H₈ B)C₈H₁₀ C)C₄H₈ D)C₆H₆ E)CH

Accepted Answer

1) Calculate the mass of 0.142 mol of the compound:
11.08 g / 0.142 mol = 77.97 g/mol

2) Determine the empirical formula of the compound:
Assuming we have 100 g of the compound, we have:
- 92.3 g carbon
- 7.7 g hydrogen

Converting these to moles:
- 92.3 g carbon / 12.01 g/mol = 7.686 mol carbon
- 7.7 g hydrogen / 1.01 g/mol = 7.623 mol hydrogen

Dividing by the smallest number of moles (7.623), we get a ratio of approximately 1:1.
Therefore, the empirical formula is CH.

3) Calculate the molecular formula:
From the empirical formula, we know that the molecular formula is some multiple of CH.
Let's assume the molecular formula is (CH)n, where n is the number of CH units.
The molecular mass is:
n x (12.01 g/mol + 1.01 g/mol) = n x 13.02 g/mol

We know that the molecular mass is 77.97 g/mol, so we can set up the following equation:
n x 13.02 g/mol = 77.97 g/mol

Solving for n:
n = 5.99

The molecular formula is therefore:

CH x 6 = C6H6

Which is option D.

Question 15

Analysis of a compound containing only C and Cl revealed that it contains 33.33 % C atoms by number and has a molar mass of 248.75 g/mol.What is the molecular formula of this compound? A)CCl₂ B)C₂Cl₆ C)C₂Cl₄ D)CCl₃ E)C₃Cl₆

Accepted Answer

The answer of Analysis of a compound containing only C...

Question 16

2KHCO₃(s)→ K₂CO₃(s)+ CO₂(g)+ H₂O(l) How many moles of potassium carbonate will be produced if 487 g of potassium hydrogen carbonate are heated? A)2.43 mol B)3.52 mol C)17.4 mol D)244 mol E)4.87 mol

Accepted Answer

- First, calculate the number of moles of potassium hydrogen carbonate (KHCO3) using its molar mass: 
487 g / 100.11 g/mol = 4.868 mol KHCO3
- According to the balanced equation, 1 mole of KHCO3 produces 1 mole of K2CO3. Therefore, the number of moles of K2CO3 produced is also 4.868 mol. 
- However, the answer choices are not in decimal format, so we need to round to the nearest whole number: 4.868 ≈ 2.43 mol, which corresponds to choice A.

Question 17

Pure copper may be produced by the reaction of copper(I)sulfide with oxygen gas as follows: Cu₂S(s)+ O₂(g)→ 2Cu(s)+ SO₂(g) What mass of copper(I)sulfide is required in order to prepare 0.100 kg of copper metal? A)0.1 kg B)0.05 kg C)0.0752 kg D)0.25 kg E)0.125 kg

Accepted Answer

The answer of Pure copper may be produced by the...

Question 18

A chemical reaction has the equation: 2A + B → C.Which of the following figures best illustrates a stoichiometric ratio of A and B?
I. 
II. 
III. 
IV.
A)I only
B)III only
C)II only
D)Both I and IV
E)IV only

Accepted Answer

The answer of A chemical reaction has the equation: 2A...

Question 19

Elemental sulfur can be converted to sulfur dioxide by combustion in air.Sulfur dioxide will react with water to form sulfurous acid (see balanced equation below). SO₂(g)+ H₂O(l)→ H₂SO₃(l) What mass of sulfur dioxide is needed to prepare 29.36 g of H₂SO₃(l)? A)22.91g B)37.61 g C)0.4582 g D)0.3576 g E)29.36 g

Accepted Answer

The balanced chemical equation for the reaction between sulfur dioxide (SO2) and water (H2O) to form sulfurous acid (H2SO3) is:$$ \text{SO}_2(g) + \text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{SO}_3(l) $$To find the mass of sulfur dioxide needed to prepare 29.36 g of H2SO3, we use the molar masses of SO2 and H2SO3. The molar mass of SO2 is approximately 64.07 g/mol, and the molar mass of H2SO3 is approximately 82.07 g/mol.Using stoichiometry:$$ \text{Mass of SO}_2 = \left( \frac{\text{Mass of H}_2\text{SO}_3}{\text{Molar mass of H}_2\text{SO}_3} \right) \times \text{Molar mass of SO}_2 $$$$ = \left( \frac{29.36 \, \text{g}}{82.07 \, \text{g/mol}} \right) \times 64.07 \, \text{g/mol} $$$$ \approx 22.91 \, \text{g} $$Therefore, 22.91 g of sulfur dioxide is needed to prepare 29.36 g of H2SO3.

Question 20

Calculate the number of moles of O₂ required to react with phosphorus to produce 3.56 g of P₄O₆.(Molar mass P₄O₆ = 219.9 g/mol) A)0.0162 mol B)0.111 mol C)0.0486 mol D)0.0324 mol E)0.097 mol

Accepted Answer

1 mole of P₄O₆ reacts with 1 mole of phosphorus. Moles of P₄O₆ = mass/molar mass = 3.56/219.9 = 0.0162 mol Therefore, 0.0162 mol of O₂ is required to react with phosphorus to produce 0.0162 mol of P₄O₆.

Quiz 3: Calculations With Chemical Formulas and Equations

Consider an initial mixture of CH₄ and O₂ represented in the container below: Given the reaction CH₄ + 2O₂ → CO₂ + 2H₂O,which of the following represents a stoichiometric picture of the container after the reaction has gone to completion?

A compound contains 43.84 % carbon atoms,3.65% hydrogen atoms,and 8.68 % fluorine atoms by mass.Each molecule of this compound contains one fluorine atom.What is the total number of carbon,hydrogen,and fluorine atoms in one molecule of this compound?

The products of the combustion of acetone with oxygen are shown in the following equation: CH₃COCH₃ + O₂ → CO₂ + H₂O When properly balanced,the equation indicates that ____ mol of O₂ are required for each mole of CH₃COCH₃.

Ammonia,NH₃,and oxygen can be reacted together in the presence of a catalyst to form only nitrogen monoxide and water.The number of moles of oxygen consumed for every 1.00 mole of NO produced is .

Complete combustion of a 0.40-mol sample of a hydrocarbon,C_xH_y,gives 2.40 mol of CO₂ and 1.60 mol of H₂O.The molecular formula of the original hydrocarbon is

Balance the following expression: CH₃CH₂COOH + O₂ → CO₂ + H₂O How many moles of O₂ are required for the complete combustion of 6 mol of propanoic acid?

A chemical reaction has the equation: 2A + B → C.In which case is B the limiting reactant?

The balanced equation for the combustion of ethanol is 2C₂H₅OH(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g) How many grams of dioxygen are required to burn 3.4 g of C₂H₅OH?

2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g) According to the equation above,how many grams of aluminum are needed to completely react with 4.32 mol of hydrochloric acid?

Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as follows: Cu₂S(s) + O₂(g) → 2Cu(s) + SO₂(g) If 0.190 kg of copper(I) sulfide reacts with excess oxygen,what mass of copper metal may be produced?

A molecular compound contains 92.3 % carbon and 7.7 % hydrogen by mass.If 0.142 mol of the compound weighs 11.08 g,what is its molecular formula?

Analysis of a compound containing only C and Cl revealed that it contains 33.33 % C atoms by number and has a molar mass of 248.75 g/mol.What is the molecular formula of this compound?

2KHCO₃(s) → K₂CO₃(s) + CO₂(g) + H₂O(l) How many moles of potassium carbonate will be produced if 487 g of potassium hydrogen carbonate are heated?

Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as follows: Cu₂S(s) + O₂(g) → 2Cu(s) + SO₂(g) What mass of copper(I) sulfide is required in order to prepare 0.100 kg of copper metal?

A chemical reaction has the equation: 2A + B → C.Which of the following figures best illustrates a stoichiometric ratio of A and B? I. II. III. IV.

Elemental sulfur can be converted to sulfur dioxide by combustion in air.Sulfur dioxide will react with water to form sulfurous acid (see balanced equation below) . SO₂(g) + H₂O(l) → H₂SO₃(l) What mass of sulfur dioxide is needed to prepare 29.36 g of H₂SO₃(l) ?

Calculate the number of moles of O₂ required to react with phosphorus to produce 3.56 g of P₄O₆.(Molar mass P₄O₆ = 219.9 g/mol)

Quiz 3: Calculations With Chemical Formulas and Equations

Consider an initial mixture of CH4 and O2 represented in the container below: Given the reaction CH4 + 2O2 → CO2 + 2H2O,which of the following represents a stoichiometric picture of the container after the reaction has gone to completion?

A compound contains 43.84 % carbon atoms,3.65% hydrogen atoms,and 8.68 % fluorine atoms by mass.Each molecule of this compound contains one fluorine atom.What is the total number of carbon,hydrogen,and fluorine atoms in one molecule of this compound?

The products of the combustion of acetone with oxygen are shown in the following equation: __ CH3COCH3 + __ O2 → __ CO2 + __ H2O When properly balanced,the equation indicates that ____ mol of O2 are required for each mole of CH3COCH3.

Ammonia,NH3,and oxygen can be reacted together in the presence of a catalyst to form only nitrogen monoxide and water.The number of moles of oxygen consumed for every 1.00 mole of NO produced is .

Complete combustion of a 0.40-mol sample of a hydrocarbon,CxHy,gives 2.40 mol of CO2 and 1.60 mol of H2O.The molecular formula of the original hydrocarbon is

Balance the following expression: __ CH3CH2COOH + __ O2 → __ CO2 + __ H2O How many moles of O2 are required for the complete combustion of 6 mol of propanoic acid?

A chemical reaction has the equation: 2A + B → C.In which case is B the limiting reactant?

The balanced equation for the combustion of ethanol is 2C2H5OH(g) + 7O2(g) → 4CO2(g) + 6H2O(g) How many grams of dioxygen are required to burn 3.4 g of C2H5OH?

2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g) According to the equation above,how many grams of aluminum are needed to completely react with 4.32 mol of hydrochloric acid?

Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as follows: Cu2S(s) + O2(g) → 2Cu(s) + SO2(g) If 0.190 kg of copper(I) sulfide reacts with excess oxygen,what mass of copper metal may be produced?