Question

21.44 moles of Si react with 17.62 moles of $N_2$. How many moles of $S{i}_3{N}_4$ will be produced?
$3Si+2N_2\to S{i}_3{N}_4$

• A. 13.87 mol
• B. 5.42 mol
• C. 10.53 mol
• D. 7.15 mol

Answer 1

The correct option is D 7.15 mol

$3Si+2N_2\to S{i}_3{N}_4$
Finding the limiting reagent:
$ForSi=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{21.44}{3}=7.15\text{For}{N}_2=\frac{17.62}{2}=8.81$
So, Si is the limiting reagent.
3 mol of Si produces 1 mole of $S{i}_3{N}_4$
$\text{21.44 moles will produce =}\frac{1}{3}\times 21.44=7.15mol\text{of}S{i}_3{N}_4$