Question

Use the chemical equation and the table to answer the question.

Pb(NO3)2(aq) + 2KBr(aq) → PbBr2(s) + 2KNO3(aq)

Reactant or Product Molar Mass (g/mol)
Pb(NO3)2 331
KBr 119
PbBr2 367
KNO3 101
When 496.5 grams of Pb(NO3)2 reacts completely with KBr, how much will the total mass of the products be?

(1 point)
Responses

853.5 g
853.5 g

702.0 g
702.0 g

496.5 g
496.5 g

550.5 g

Answer 1

To determine the total mass of the products when 496.5 grams of Pb(NO3)2 reacts completely with KBr, we can use the principle of conservation of mass. According to this principle, the total mass of the reactants equals the total mass of the products in a chemical reaction.

1. **Calculate the mass of the reactants:**
\[
\text{Mass of Pb(NO3)2} = 496.5 \text{ g}
\]

The mass of KBr will be determined based on the stoichiometry of the reaction with Pb(NO3)2.

According to the balanced equation:
\[
\text{1 mole of } Pb(NO3)2 \text{ reacts with } 2 \text{ moles of } KBr
\]

First, let's calculate the number of moles of Pb(NO3)2:
\[
\text{Number of moles of } Pb(NO3)2 = \frac{\text{mass}}{\text{molar mass}} = \frac{496.5 \text{ g}}{331 \text{ g/mol}} \approx 1.50 \text{ moles}
\]

Since 2 moles of KBr are required for every mole of Pb(NO3)2, we need:
\[
\text{Number of moles of KBr} = 2 \times 1.50 \text{ moles} = 3.00 \text{ moles}
\]

Now calculate the mass of KBr required:
\[
\text{Mass of KBr} = \text{number of moles} \times \text{molar mass} = 3.00 \text{ moles} \times 119 \text{ g/mol} = 357.0 \text{ g}
\]

2. **Calculate the total mass of the reactants:**
\[
\text{Total mass of reactants} = \text{mass of Pb(NO3)2} + \text{mass of KBr} = 496.5 \text{ g} + 357.0 \text{ g} = 853.5 \text{ g}
\]

3. **Determine the total mass of the products:**
By the conservation of mass, the mass of the products will equal the mass of the reactants. Therefore, the total mass of the products will also be:
\[
\text{Total mass of products} = 853.5 \text{ g}
\]

So, the answer is **853.5 g**.