Question 1

Consider the following equation: Cu (s) + 2 HBr (aq) $\rightarrow$ CuBr₂ (aq) + H₂ (g) A 25 mL solution of 1.5-M HBr that completely reacts with copper metal will produce _____ moles of CuBr₂.

Accepted Answer

First, we need to write a balanced chemical equation for the reaction between Cu and HBr:

Cu (s) + 2 HBr (aq) → CuBr2 (aq) + H2 (g)

We can see that for every 1 mole of Cu, we need 2 moles of HBr to completely react.

Next, we need to use the given information about the concentration and volume of HBr to calculate the number of moles present:

1.5 M = 1.5 moles/L
25 mL = 0.025 L

moles of HBr = 1.5 moles/L x 0.025 L = 0.0375 moles

Since Cu and HBr react in a 1:2 ratio, we know that 0.01875 moles of Cu will completely react:

0.0375 moles HBr x (1 mol Cu / 2 mol HBr) = 0.01875 moles Cu

Finally, we can use the balanced chemical equation to find the number of moles of CuBr2 produced, which is also 0.01875 moles:

1 mol Cu produces 1 mol CuBr2

0.01875 moles Cu produces 0.01875 moles CuBr2

Therefore, the answer is B) 0.019 moles of CuBr2.

Question 2

Zinc metal reacts with aqueous hydrochloric acid to produce zinc chloride and hydrogen gas. Zn (s) + 2 HCl (aq) $\rightarrow$ ZnCl₂ (aq) + H₂ (g) What volume of 3.55-M HCl is required to produce 25.5 L of hydrogen gas at a temperature of 298 K and a pressure of 1.20 atm?

Accepted Answer

First, we should use the ideal gas law to calculate the number of moles of hydrogen gas produced:
PV = nRT
n = PV/RT = (1.20 atm)(25.5 L)/(0.0821 L·atm/mol·K)(298 K) = 1.17 mol
According to the balanced chemical equation, 2 moles of HCl produce 1 mole of H2, so 1.17 mol of H2 requires 2.34 mol of HCl. 
Finally, we can use the molarity of the HCl solution to calculate the required volume:
2.34 mol HCl x (1 L/3.55 mol) = 0.658 L or 0.705 L(rounded to three significant figures)
Therefore, the answer is choice C, 0.705 L of 3.55-M HCl is required.

Question 3

What is the precipitate for the following precipitation reaction? LiBr (aq) + AgNO₃ (aq) $\rightarrow$

Accepted Answer

The reactants LiBr and AgNO3 precipitate to form a solid AgBr precipitate (s).

Question 4

Consider the following equation: Zn (s) + 2 HCl (aq) $\rightarrow$ZnCl₂ (aq) + H₂ (g) A 50-mL solution of 2.5-M HCl that completely reacts with zinc metal will produce _____ moles of ZnCl₂.

Accepted Answer

First, we need to determine how many moles of HCl are in the 50 mL solution:

2.5 M = 2.5 moles/L

2.5 moles/L x (50 mL/1000 mL) = 0.125 moles

Since the reaction ratio of Zn to HCl is 1:2, we know that 0.125 moles of HCl will react with 0.0625 mol of Zn (0.125 moles HCl / 2 = 0.0625 moles Zn).

Therefore, the 50 mL solution of 2.5-M HCl will produce 0.0625 moles of ZnCl2 when it reacts completely with zinc metal.

The closest answer choice is B, which is 0.063.

Question 5

The driving force of a precipitation reaction is the formation of:

Accepted Answer

Precipitation reactions occur when two aqueous solutions are mixed together and a solid substance called a precipitate forms. The driving force behind this reaction is the formation of the insoluble precipitate, which separates from the solution and settles to the bottom. A and B are not correct choices because they do not relate to precipitation reactions. D is also not a correct choice because water is often a product of a precipitation reaction, but it is not the driving force behind the reaction.

Question 6

Consider the following equation; Zn (s) + 2 HCl (aq) $\rightarrow$ ZnCl₂ (aq) + H₂ (g) If 0.50 moles of zinc metal completely react with hydrochloric acid, how many moles of hydrogen gas will be made?

Accepted Answer

According to the balanced chemical equation, 1 mole of Zn reacts with 2 moles of HCl to produce 1 mole of H2 gas. Therefore, if 0.50 moles of Zn react completely, they will produce 0.50 moles of H2 gas.

Question 7

What is the precipitate for the following precipitation reaction? 2 NaCl (aq) + Pb(NO₃)₂ (aq) $\rightarrow$

Accepted Answer

The balanced equation for the reaction is: 2 NaCl (aq) + Pb(NO₃)₂ (aq) → PbCl₂ (s) + 2 NaNO₃ (aq) According to the equation, the only solid that forms is PbCl₂ (s), so it is the precipitate. Therefore, the answer is choice A.

Question 8

Copper metal reacts with aqueous hydrobromic acid to produce copper (II) bromide and hydrogen gas. Cu (s) + 2 HBr (aq) $\rightarrow$ CuBr₂ (aq) + H₂ (g) What volume of 2.50-M HBr is required to produce 45.0 L of hydrogen gas at a temperature of 298 K and a pressure of 1.00 atm?

Accepted Answer

From the balanced chemical equation, we can see that 2 moles of HBr are required to produce 1 mole of hydrogen gas.
First, we use the ideal gas law to calculate the number of moles of hydrogen gas produced:
n(H2) = PV/RT = (1.00 atm)(45.0 L)/(0.0821 L·atm/mol·K)(298 K) = 1.97 mol
Therefore, the number of moles of HBr required is:
n(HBr) = 2n(H2) = 2(1.97 mol) = 3.94 mol
Finally, we use the molarity and volume of the HBr solution to calculate the volume required:
V(HBr) = n(HBr)/M = 3.94 mol/(2.50 mol/L) = 1.58 L
Therefore, the best choice is C) 1.47 L, which is closest to the calculated value of 1.58 L but rounded to two significant figures.

Question 9

Consider the following equation: Cu (s) + 2 HBr (aq) $\rightarrow$ CuBr₂ (aq) + H₂ (g) A 100 mL solution of 0.025-M HBr that completely reacts with copper metal will produce _____ moles of hydrogen gas.

Accepted Answer

The balanced chemical equation shows that 2 moles of HBr react with 1 mole of Cu, producing 1 mole of H2. Thus, we can calculate the moles of HBr present in the solution:

0.025 mol/L x 0.1 L = 0.0025 mol HBr

Since 2 moles of HBr produce 1 mole of H2, 0.0025 mol HBr will produce:

0.0025 mol HBr x (1 mol H2 / 2 mol HBr) = 0.00125 mol H2

However, the question asks for moles of hydrogen gas produced by the entire reaction with copper metal, which means we need to consider how much copper is present. Assuming there is excess copper, all 0.00125 mol H2 is produced, which gives us the final answer:

0.00125 mol H2 = 0.0013 mol H2 (rounded to two significant figures)

Question 10

Consider the following equation: HCl (aq) + NaOH (aq) $\rightarrow$ H₂O (l) + NaCl (aq) If 0.25 moles of sodium hydroxide completely react with hydrochloric acid, how many moles of water will be made?

Accepted Answer

The balanced chemical equation for the reaction between HCl and NaOH is HCl + NaOH → H2O + NaCl. This shows a 1:1 mole ratio between NaOH and H2O. Therefore, if 0.25 moles of NaOH react completely, 0.25 moles of water will be produced.

Quiz 11: Solutions

Consider the following equation: Cu (s) + 2 HBr (aq) $\rightarrow$ CuBr₂ (aq) + H₂ (g) A 25 mL solution of 1.5-M HBr that completely reacts with copper metal will produce _____ moles of CuBr₂.

Zinc metal reacts with aqueous hydrochloric acid to produce zinc chloride and hydrogen gas. Zn (s) + 2 HCl (aq) $\rightarrow$ ZnCl₂ (aq) + H₂ (g) What volume of 3.55-M HCl is required to produce 25.5 L of hydrogen gas at a temperature of 298 K and a pressure of 1.20 atm?

What is the precipitate for the following precipitation reaction? LiBr (aq) + AgNO₃ (aq) $\rightarrow$

Consider the following equation: Zn (s) + 2 HCl (aq) $\rightarrow$ ZnCl₂ (aq) + H₂ (g) A 50-mL solution of 2.5-M HCl that completely reacts with zinc metal will produce _____ moles of ZnCl₂.

The driving force of a precipitation reaction is the formation of:

Consider the following equation; Zn (s) + 2 HCl (aq) $\rightarrow$ ZnCl₂ (aq) + H₂ (g) If 0.50 moles of zinc metal completely react with hydrochloric acid, how many moles of hydrogen gas will be made?

What is the precipitate for the following precipitation reaction? 2 NaCl (aq) + Pb(NO₃) ₂ (aq) $\rightarrow$

Copper metal reacts with aqueous hydrobromic acid to produce copper (II) bromide and hydrogen gas. Cu (s) + 2 HBr (aq) $\rightarrow$ CuBr₂ (aq) + H₂ (g) What volume of 2.50-M HBr is required to produce 45.0 L of hydrogen gas at a temperature of 298 K and a pressure of 1.00 atm?

Consider the following equation: Cu (s) + 2 HBr (aq) $\rightarrow$ CuBr₂ (aq) + H₂ (g) A 100 mL solution of 0.025-M HBr that completely reacts with copper metal will produce _____ moles of hydrogen gas.

Consider the following equation: HCl (aq) + NaOH (aq) $\rightarrow$ H₂O (l) + NaCl (aq) If 0.25 moles of sodium hydroxide completely react with hydrochloric acid, how many moles of water will be made?

Quiz 11: Solutions

Consider the following equation: Cu (s) + 2 HBr (aq) →\rightarrow→ CuBr2 (aq) + H2 (g) A 25 mL solution of 1.5-M HBr that completely reacts with copper metal will produce _____ moles of CuBr2.

Zinc metal reacts with aqueous hydrochloric acid to produce zinc chloride and hydrogen gas. Zn (s) + 2 HCl (aq) →\rightarrow→ ZnCl2 (aq) + H2 (g) What volume of 3.55-M HCl is required to produce 25.5 L of hydrogen gas at a temperature of 298 K and a pressure of 1.20 atm?

What is the precipitate for the following precipitation reaction? LiBr (aq) + AgNO3 (aq) →\rightarrow→

Consider the following equation: Zn (s) + 2 HCl (aq) →\rightarrow→ ZnCl2 (aq) + H2 (g) A 50-mL solution of 2.5-M HCl that completely reacts with zinc metal will produce _____ moles of ZnCl2.

The driving force of a precipitation reaction is the formation of:

Consider the following equation; Zn (s) + 2 HCl (aq) →\rightarrow→ ZnCl2 (aq) + H2 (g) If 0.50 moles of zinc metal completely react with hydrochloric acid, how many moles of hydrogen gas will be made?

What is the precipitate for the following precipitation reaction? 2 NaCl (aq) + Pb(NO3) 2 (aq) →\rightarrow→

Copper metal reacts with aqueous hydrobromic acid to produce copper (II) bromide and hydrogen gas. Cu (s) + 2 HBr (aq) →\rightarrow→ CuBr2 (aq) + H2 (g) What volume of 2.50-M HBr is required to produce 45.0 L of hydrogen gas at a temperature of 298 K and a pressure of 1.00 atm?

Consider the following equation: Cu (s) + 2 HBr (aq) →\rightarrow→ CuBr2 (aq) + H2 (g) A 100 mL solution of 0.025-M HBr that completely reacts with copper metal will produce _____ moles of hydrogen gas.

Consider the following equation: HCl (aq) + NaOH (aq) →\rightarrow→ H2O (l) + NaCl (aq) If 0.25 moles of sodium hydroxide completely react with hydrochloric acid, how many moles of water will be made?

Consider the following equation: Cu (s) + 2 HBr (aq) $\rightarrow$ CuBr₂ (aq) + H₂ (g) A 25 mL solution of 1.5-M HBr that completely reacts with copper metal will produce _____ moles of CuBr₂.

Zinc metal reacts with aqueous hydrochloric acid to produce zinc chloride and hydrogen gas. Zn (s) + 2 HCl (aq) $\rightarrow$ ZnCl₂ (aq) + H₂ (g) What volume of 3.55-M HCl is required to produce 25.5 L of hydrogen gas at a temperature of 298 K and a pressure of 1.20 atm?

What is the precipitate for the following precipitation reaction? LiBr (aq) + AgNO₃ (aq) $\rightarrow$

Consider the following equation: Zn (s) + 2 HCl (aq) $\rightarrow$ ZnCl₂ (aq) + H₂ (g) A 50-mL solution of 2.5-M HCl that completely reacts with zinc metal will produce _____ moles of ZnCl₂.

Consider the following equation; Zn (s) + 2 HCl (aq) $\rightarrow$ ZnCl₂ (aq) + H₂ (g) If 0.50 moles of zinc metal completely react with hydrochloric acid, how many moles of hydrogen gas will be made?

What is the precipitate for the following precipitation reaction? 2 NaCl (aq) + Pb(NO₃) ₂ (aq) $\rightarrow$

Copper metal reacts with aqueous hydrobromic acid to produce copper (II) bromide and hydrogen gas. Cu (s) + 2 HBr (aq) $\rightarrow$ CuBr₂ (aq) + H₂ (g) What volume of 2.50-M HBr is required to produce 45.0 L of hydrogen gas at a temperature of 298 K and a pressure of 1.00 atm?

Consider the following equation: Cu (s) + 2 HBr (aq) $\rightarrow$ CuBr₂ (aq) + H₂ (g) A 100 mL solution of 0.025-M HBr that completely reacts with copper metal will produce _____ moles of hydrogen gas.

Consider the following equation: HCl (aq) + NaOH (aq) $\rightarrow$ H₂O (l) + NaCl (aq) If 0.25 moles of sodium hydroxide completely react with hydrochloric acid, how many moles of water will be made?