Sure, let's solve this step by step:
1. Identify the Molar Masses:
- The molar mass of silver nitrate (AgNO₃) is 169.87 g/mol.
- The molar mass of sodium chromate (Na₂CrO₄) is 161.97 g/mol.
2. Calculate the Moles of Silver Nitrate:
- Given mass of AgNO₃ is 44.7 grams.
- To find the moles of AgNO₃, use the formula:
[tex]\[
\text{moles of AgNO₃} = \frac{\text{mass of AgNO₃}}{\text{molar mass of AgNO₃}}
\][/tex]
[tex]\[
\text{moles of AgNO₃} = \frac{44.7 \text{ grams}}{169.87 \text{ g/mol}} \approx 0.2631 \text{ moles}
\][/tex]
3. Determine the Moles of Sodium Chromate Needed:
- From the balanced equation [tex]\(2 \text{AgNO₃} + \text{Na₂CrO₄} \rightarrow \text{Ag₂CrO₄} + 2\text{NaNO₃}\)[/tex], the molar ratio of AgNO₃ to Na₂CrO₄ is 2:1.
- Therefore, the moles of Na₂CrO₄ needed will be half the moles of AgNO₃:
[tex]\[
\text{moles of Na₂CrO₄} = \frac{\text{moles of AgNO₃}}{2}
\][/tex]
[tex]\[
\text{moles of Na₂CrO₄} = \frac{0.2631 \text{ moles}}{2} \approx 0.1316 \text{ moles}
\][/tex]
4. Calculate the Mass of Sodium Chromate Needed:
- Use the moles of Na₂CrO₄ and its molar mass to find the mass needed:
[tex]\[
\text{mass of Na₂CrO₄} = \text{moles of Na₂CrO₄} \times \text{molar mass of Na₂CrO₄}
\][/tex]
[tex]\[
\text{mass of Na₂CrO₄} = 0.1316 \text{ moles} \times 161.97 \text{ g/mol} \approx 21.31 \text{ grams}
\][/tex]
Therefore, to react completely with 44.7 grams of silver nitrate (AgNO₃), you would need approximately 21.31 grams of sodium chromate (Na₂CrO₄).
