To determine how many grams of KNO₃ can form, we proceed by following the stoichiometric relationships given in the balanced chemical equation:
[tex]\[ 2 \, \text{AgNO}_3 (aq) + \text{K}_2\text{CrO}_4 (aq) \rightarrow \text{Ag}_2\text{CrO}_4 (s) + 2 \, \text{KNO}_3 (aq) \][/tex]
1. Identify the molar ratios from the balanced equation:
From the balanced equation, we see that 1 mole of K₂CrO₄ reacts to produce 2 moles of KNO₃.
2. Determine the moles of K₂CrO₄ provided:
From the problem, we know that 0.040 mol of K₂CrO₄ react in the solution.
3. Calculate the moles of KNO₃ produced:
Using the stoichiometric ratio (1 mole of K₂CrO₄ produces 2 moles of KNO₃), we find:
[tex]\[
\text{Moles of KNO₃} = 2 \times 0.040 \, \text{mol} = 0.080 \, \text{mol}
\][/tex]
4. Calculate the mass of KNO₃ formed:
The molar mass of KNO₃ is given as 101.11 g/mol. To find the mass, we use the formula:
[tex]\[
\text{Mass of KNO₃} = \text{moles of KNO₃} \times \text{molar mass of KNO₃}
\][/tex]
Substituting the known values, we get:
[tex]\[
\text{Mass of KNO₃} = 0.080 \, \text{mol} \times 101.11 \, \text{g/mol} = 8.0888 \, \text{g}
\][/tex]
Therefore, the mass of KNO₃ that can form during the reaction is 8.0888 grams.
