2Cr+2H3PO4=>2CrPO4+3H2 | You are given 40g of Cr and 90g of H3PO4. If CrPO4 is the target product, which reactant is limiting? If 60g of CrPO4 is actually produced, what is the percent yield?

Answer 1

2Cr + 2H₃PO₄ ==> 2CrPO₄ + 3H₂

moles Cr present = 40 g x 1 mol/52 g = 0.769 moles Cr

moles H₃PO₄ = 90 g x 1 mol/98 g = 0.918 moles H₃PO₄

Limiting reactant = Cr because the mole ratio is 1:1 and there are fewer moles of Cr than H₃PO₄

Theoretical yield = 0.769 moles Cr x 2 moles CrPO₄/2 moles Cr x 147 g/mole = 113 g

% yield = actual yield/theoretical yield (x100) = 60 g/113 g (x100) = 53% yield (2 sig. figs.)