Question

Silicon nitride ($S{i}_3{N}_4$) a valuable ceramic is made by the direct combination of silicon and nitrogen at high temperature. How much silicon must react with excess nitrogen to prepare$125g$silicon nitride if the yield of the reaction is$95.0\%$?

Short answer

Amount if silicon required from the reaction is 79.3g.

Step-by-step solution

Balanced chemical equation

The silicon nitride made from direct combination of silicon and nitrogen.

The balanced reaction for this is:

$Si+{\mathrm{N}}_2\to S{i}_3{N}_2$

Assigning coefficient 1to $S{i}_3{N}_4$, so left side coefficient of si must be 3 to balance amount of si on both sides.

Same argument foratoms, coefficient ofmust be.

So balanced equation is:

$\boxed{3Si+2N_2\to S{i}_3{N}_4}$

Theoretical yield of

$Percentageyield=\frac{ActualyieldofS{i}_3{N}_4}{TheoriticalyieldofS{i}_3{N}_4}\times 100$

Given that, percentage yield is95.0% , actual yield of${Si}_3{N}_4$is:

$\frac{125gS{i}_3{N}_4}{Theoriticalyield}\times 100=95$

$Theoriticalyield=\frac{125gS{i}_3{N}_4}{95}\times 100$

$=132gS{i}_3{N}_4$

Required amount of silicon

According to balanced equation, $1molS{i}_3{N}_4$is produced from $3\mathrm{mol}\mathrm{Si}$. Then amount if silicon required to produce is $132gS{i}_3{N}_4$

$132gS{i}_3{N}_4\times \frac{1molS{i}_3{N}_4}{140molS{i}_3{N}_4}\times \frac{3molSi}{1molS{i}_3{N}_4}\times \frac{28.09gmolSi}{1molSi}$

Hence, the amount of silicon required for the reaction is 79.3g