Question

practice iii
answer the question below using the following
balanced equation:
3 si(s) + 2 n₂(g) → 1 si₃n₄(s)
since si is the limiting reactant, it will be
completely used up by the equation. however,
n₂, the excess reactant, would not be
completely used up.
now lets say we have:
16.24 moles of si
15.38 moles of n₂
how many moles of n₂ would be
left over after the reaction is
complete?
moles n2

Explanation:

Step1: Calculate moles of \(N_2\) used

From the balanced equation \(3Si_{(s)}+2N_{2(g)}\to Si_3N_{4(s)}\), the mole ratio of \(Si\) to \(N_2\) is \(3:2\).
If there are \(n(Si) = 16.24\) moles of \(Si\), let \(n_{used}(N_2)\) be the moles of \(N_2\) used.
Using the ratio \(\frac{n(Si)}{n_{used}(N_2)}=\frac{3}{2}\), then \(n_{used}(N_2)=\frac{2\times n(Si)}{3}\)
Substitute \(n(Si) = 16.24\) moles: \(n_{used}(N_2)=\frac{2\times16.24}{3}=\frac{32.48}{3}\approx10.83\) moles

Step2: Calculate moles of \(N_2\) left

We know that \(n_{initial}(N_2)=15.38\) moles.
Using the formula \(n_{left}(N_2)=n_{initial}(N_2)-n_{used}(N_2)\)
Substitute \(n_{initial}(N_2) = 15.38\) moles and \(n_{used}(N_2)\approx10.83\) moles
\(n_{left}(N_2)=15.38 - 10.83=4.55\) moles

Answer:

\(4.55\) moles