Question

1. silicon nitride is a very hard, high - temperature - resistant ceramic used as a component of turbine blades in jet engines. it is prepared by the reaction of nitrogen gas on silicon solid under special conditions.

a. balance the chemical equation
b. which is the limiting reactant when 2.00 g of si and 1.50 g of n₂ react?
c. the production has a percent yield of 48%. calculate the mass of silicon nitride produce.

Explanation:

Part a

Step1: Write the unbalanced equation

The reaction is \(Si + N_2
ightarrow Si_3N_4\)

Step2: Balance Si atoms

There are 3 Si atoms in \(Si_3N_4\). So we put a coefficient of 3 in front of \(Si\): \(3Si+N_2
ightarrow Si_3N_4\)

Step3: Balance N atoms

There are 4 N atoms in \(Si_3N_4\). So we put a coefficient of 2 in front of \(N_2\): \(3Si + 2N_2
ightarrow Si_3N_4\)

Step1: Calculate the molar mass

The molar mass of \(Si\) is \(M_{Si}=28.09\space g/mol\), and the molar mass of \(N_2\) is \(M_{N_2} = 28.02\space g/mol\)

Step2: Calculate the number of moles

For \(Si\): \(n_{Si}=\frac{m_{Si}}{M_{Si}}=\frac{2.00\space g}{28.09\space g/mol}\approx0.0712\space mol\)
For \(N_2\): \(n_{N_2}=\frac{m_{N_2}}{M_{N_2}}=\frac{1.50\space g}{28.02\space g/mol}\approx0.0535\space mol\)

Step3: Use the stoichiometric ratio

From the balanced equation \(3Si + 2N_2
ightarrow Si_3N_4\), the ratio \(n_{Si}/n_{N_2}\) (theoretical) is \(3/2 = 1.5\)
The ratio \(n_{Si}/n_{N_2}\) (experimental) is \(\frac{0.0712}{0.0535}\approx1.33\)
Since \(1.33<1.5\), \(N_2\) is in excess and \(Si\) is the limiting reactant.

Step1: Calculate the theoretical yield

From the balanced equation, \(3\space mol\space Si\) produces \(1\space mol\space Si_3N_4\)
\(n_{Si}=0.0712\space mol\), so \(n_{Si_3N_4}^{theo}=\frac{0.0712}{3}\space mol\)
The molar mass of \(Si_3N_4\) is \(M_{Si_3N_4}=3\times28.09 + 4\times14.01=140.31\space g/mol\)
\(m_{Si_3N_4}^{theo}=n_{Si_3N_4}^{theo}\times M_{Si_3N_4}=\frac{0.0712}{3}\times140.31\space g\approx3.33\space g\)

Step2: Calculate the actual yield

Percent yield \(=\frac{m_{actual}}{m_{theo}}\times100\%\)
\(m_{actual}=m_{theo}\times\frac{\text{Percent yield}}{100\%}\)
\(m_{actual}=3.33\times\frac{48}{100}\space g = 1.60\space g\)

Answer:

\(3Si + 2N_2
ightarrow Si_3N_4\)

Part b