chemistry- limiting reagent?

Question

i need tp produce 11.3 kg of sodium sulfide frem sodium metal and dihydrogen sulfide. how many grams of starting material do I need if I know this reaction proceeds in 65% yield?

Answer 1

11.3 kg is equivalent to 65% yield
17.4 kg is equivalent to 100% yield.
17.4 kg = 17400 g (NaS)
Mr(Na2S) = 23 + 23 + 32 = 78
Moles (Na2S) = 17400/78 = 223
Reaction formula
2Na + H2S = Na2S + H2
The molar ratios are :-
Na: Na2S :: 2 : 1
So to form one mole of Na2S two moles of Na are needed.
So 223 moles(Na2S) are formed from 446 moles(Na)
Therefore mass (Na) = 446 x 23 = 10258 grams (10.258 Kg).
So you need 10.258 kg of sodium.