Silver(I) nitrate and sodium sulfate react to form silver(I) sulfate and sodium nitrate.
Suppose 45.98 mL of 0.060 M silver(I) nitrate solution and 26.64 mL of 0.10 M sodium sulfate solution are mixed. What is the maximum number of grams of silver(I) sulfate that can be made?
This is a limiting reagent problem and it's almost like the one I worked for you earlier (2-3 hours ago). Refer to that. moles = M x L
To find the maximum number of grams of silver(I) sulfate that can be made, we need to determine the limiting reagent in the reaction. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
First, we need to calculate the number of moles of silver(I) nitrate and sodium sulfate:
Number of moles of silver(I) nitrate = volume of silver(I) nitrate solution (in L) × molarity of silver(I) nitrate
Number of moles of silver(I) nitrate = 0.04598 L × 0.060 mol/L = 0.0027588 mol
Number of moles of sodium sulfate = volume of sodium sulfate solution (in L) × molarity of sodium sulfate
Number of moles of sodium sulfate = 0.02664 L × 0.10 mol/L = 0.002664 mol
Based on the balanced chemical equation, the ratio of moles of silver(I) nitrate to silver(I) sulfate is 1:1. Therefore, the number of moles of silver(I) sulfate formed will be the same as the number of moles of silver(I) nitrate.
Now, we need to calculate the number of moles of silver(I) sulfate:
Number of moles of silver(I) sulfate = number of moles of silver(I) nitrate = 0.0027588 mol
Finally, we can calculate the mass of silver(I) sulfate using its molar mass:
Mass of silver(I) sulfate = number of moles of silver(I) sulfate × molar mass of silver(I) sulfate
Mass of silver(I) sulfate = 0.0027588 mol × (107.87 g/mol) = 0.2975 g
Therefore, the maximum number of grams of silver(I) sulfate that can be made is approximately 0.2975 grams.