Question

when 79.0 g of calcium and 54.7 g of nitrogen gas undergo a reaction that has a 90.9% yield, what mass of calcium nitride forms?

__________g calcium nitride

Answer 1

3Ca + N₂ ------------> Ca₃N₂

Now, molar mass of Ca = 40 g/mole

molar mass of N₂ = 28 g/mole

moles of 79 g of Ca = 79/40 = 1.975

moles of 54.7 g of N₂ = 54.7/28 = 1.954

Now, as per the balanced reaction:- Ca and N₂ reacts in the ratio of 3:1

Thus, for 1.954 moles of N₂ , 5.862 moles of Ca is required

Clearly, Ca is the limiting reagent

Thus, theoretical yield of  Ca₃N₂ = (1/3)*moles of Ca = (1/3)*1.975 = 0.6583 moles

Theoretical yield of  Ca₃N₂ in grams = 0.6853*148 = 97.43 g

Actual yield of  Ca₃N₂ = 0.909*97.43 = 88.57 g