To find the total mass of the products when 496.5 grams of Pb(NO₃)₂ reacts completely with KBr, we first need to write the balanced chemical equation for the reaction.
The reaction between lead(II) nitrate and potassium bromide can be written as:
\[ \text{Pb(NO}_3\text{)}_2 (aq) + 2 \text{KBr} (aq) \rightarrow \text{PbBr}_2 (s) + 2 \text{KNO}_3 (aq) \]
This equation shows that 1 mole of lead(II) nitrate reacts with 2 moles of potassium bromide to produce 1 mole of lead(II) bromide and 2 moles of potassium nitrate.
Now, we'll need to determine the molar masses of the reactants involved in the reaction:
- Molar mass of \( \text{Pb(NO}_3\text{)}_2 \):
- Pb: 207.2 g/mol
- N: 14.01 g/mol (2 atoms)
- O: 16.00 g/mol (6 atoms)
\[ \text{Molar mass of Pb(NO}_3\text{)}_2 = 207.2 + (2 \times 14.01) + (6 \times 16.00) = 207.2 + 28.02 + 96.00 = 331.22 , \text{g/mol} \]
- Molar mass of \( \text{KBr} \):
- K: 39.10 g/mol
- Br: 79.90 g/mol
\[ \text{Molar mass of KBr } = 39.10 + 79.90 = 119.00 , \text{g/mol} \]
Now, we find the moles of \( \text{Pb(NO}_3\text{)}_2 \) present in 496.5 grams:
\[ \text{Moles of } \text{Pb(NO}_3\text{)}_2 = \frac{496.5 , \text{g}}{331.22 , \text{g/mol}} \approx 1.5 , \text{mol} \]
From the balanced equation, we see that 1 mole of \( \text{Pb(NO}_3\text{)}_2 \) reacts with 2 moles of \( \text{KBr} \). Therefore, the moles of KBr required is:
\[ \text{Moles of KBr} = 2 \times 1.5 , \text{mol} = 3 , \text{mol} \]
Now, we find the mass of KBr needed:
\[ \text{Mass of KBr} = 3 , \text{mol} \times 119.00 , \text{g/mol} = 357.00 , \text{g} \]
Next, let's summarize the reaction substances:
- Starting materials:
- Lead(II) nitrate: 496.5 g
- Potassium bromide: 357.0 g
- Total mass of reactants = 496.5 g + 357.0 g = 853.5 g
According to the law of conservation of mass, the total mass of the products will equal the total mass of the reactants. Therefore, the total mass of the products will also be:
\[ \text{Total mass of products} = 853.5 , \text{grams} \]
So, the total mass of the products will be 853.5 grams.
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