Silver(I) nitrate and sodium sulfate react to form silver(I) sulfate and sodium nitrate.
Suppose 49.02 mL of 0.040 M silver(I) nitrate solution and 39.48 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?
Bob Pursley worked a similar problem several years ago. Use this as an example.
"moles silver: .0435*.04
moles sulfate:.02658*.1
To make Ag2SO4, from the above, one is limiting. So of the two inputs above, which will be limiting?
chem."
To find the maximum number of grams of silver(I) sulfate that can be made, we need to determine which reactant is limiting and calculate the amount of silver(I) sulfate formed from the limiting reactant.
First, we need to find the moles of each reactant. We can use the formula:
moles = concentration (M) x volume (L)
For silver(I) nitrate:
moles of silver(I) nitrate = 0.040 M x 0.04902 L = 0.0019608 mol
For sodium sulfate:
moles of sodium sulfate = 0.10 M x 0.03948 L = 0.003948 mol
Next, we need to determine which reactant is limiting. The limiting reactant is the one that is completely consumed and restricts the amount of product formed.
We can compare the moles of silver(I) nitrate and sodium sulfate based on their stoichiometric coefficients in the balanced chemical equation.
From the balanced equation, we see that the stoichiometric ratio between silver(I) nitrate and silver(I) sulfate is 1:1. This means that for every 1 mole of silver(I) nitrate, we get 1 mole of silver(I) sulfate.
From the stoichiometry, we can calculate the moles of silver(I) sulfate that can be formed from the available moles of silver(I) nitrate:
moles of silver(I) sulfate from silver(I) nitrate = 0.0019608 mol
Similarly, from the balanced equation, we see that the stoichiometric ratio between sodium sulfate and silver(I) sulfate is 1:1. This means that for every 1 mole of sodium sulfate, we get 1 mole of silver(I) sulfate.
From the stoichiometry, we can calculate the moles of silver(I) sulfate that can be formed from the available moles of sodium sulfate:
moles of silver(I) sulfate from sodium sulfate = 0.003948 mol
Since the stoichiometric ratio between the two reactants is 1:1 for the formation of silver(I) sulfate, the limiting reactant is the one that produces the smaller amount of moles. That is, silver(I) nitrate is the limiting reactant in this case.
To calculate the mass of silver(I) sulfate formed, we need to use the molar mass of silver(I) sulfate, which is 311.8 g/mol.
mass of silver(I) sulfate = moles of silver(I) sulfate x molar mass of silver(I) sulfate
mass of silver(I) sulfate = 0.0019608 mol x 311.8 g/mol = 0.611 g
Therefore, the maximum number of grams of silver(I) sulfate that can be made is 0.611 grams.