How many moles MgIn2S4 could you make from 1.00 g of Mg , 1.00 g of In and 1.00 g of S ?

Question

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Answer 1

Moles Mg = 1.00 g/ 24.312 = 0.0411
Moles In = 1.00 g/ 114.82 = 0.00871
Moles S = 1.00 g / 32.064 = 0.0312
The ratio is 1 : 2 : 4
In is the limiting reactant
Mole MgIn2S4 = 0.00871 / 2 = 0.00436

Answer 2

Atomic weights: Mg=24 In=115 S=32

1.00gMg x 1molMg/24g = 0.0417mol Mg

1.00gIn x 1molIn/115gIn = 0.00870mol In

1.00gS x 1molS/32gS = 0.0312mol S

0.0417 mole Mg needs 4 x 0.0417 = 0.1668 moles S. You haven't got that much S. 0.00870 mole In needs 2 x 0.00870 = 0.0174 mol S. So In is the limiting reagent. Going back to In:

0.00870molIn x 1molMhIn2S4/2molIn = 0.00435 mole MgIn2S4