Question

elemental silicon can be produced by the reduction of silicon dioxide, or sand, with carbon:

sio₂ + 2c → si + 2co

when 35.0 kg of sio₂ react with 25.3 kg of c, 14.4 kg of si are recovered. what is the percent yield for the reaction?

% yield =

Explanation:

Step1: Calculate the molar mass

The molar mass of \(SiO_{2}\) is \(M_{SiO_{2}}=(28.09 + 2\times16.00)\space g/mol=60.09\space g/mol\).
The molar mass of \(C\) is \(M_{C} = 12.01\space g/mol\).
The molar mass of \(Si\) is \(M_{Si}=28.09\space g/mol\).
Convert the mass of reactants to moles:
\(n_{SiO_{2}}=\frac{35.0\times10^{3}\space g}{60.09\space g/mol}\approx582.5\space mol\)
\(n_{C}=\frac{25.3\times10^{3}\space g}{12.01\space g/mol}\approx2106.6\space mol\)

Step2: Determine the limiting reactant

From the balanced equation \(SiO_{2}+2C
ightarrow Si + 2CO\), the mole ratio of \(SiO_{2}\) to \(C\) is \(1:2\).
For \(n_{SiO_{2}} = 582.5\space mol\), the required \(n_{C}\) is \(2\times582.5\space mol = 1165\space mol\).
Since \(n_{C}(2106.6\space mol)>1165\space mol\), \(SiO_{2}\) is the limiting reactant.

Step3: Calculate the theoretical yield of \(Si\)

From the balanced equation, \(n_{Si}=n_{SiO_{2}}\) (mole ratio \(1:1\)).
The theoretical mass of \(Si\), \(m_{theo}=n_{Si}\times M_{Si}\)
Since \(n_{Si}=n_{SiO_{2}} = 582.5\space mol\), \(m_{theo}=582.5\space mol\times28.09\space g/mol\)
\(m_{theo}=582.5\times28.09\space g\approx16362.4\space g = 16.3624\space kg\)

Step4: Calculate the percent yield

Percent yield formula: \(\%\text{yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times100\%\)
Given actual yield \(m_{actual}=14.4\space kg\)
\(\%\text{yield}=\frac{14.4\space kg}{16.3624\space kg}\times100\%\approx88.0\%\)

Answer:

\(88.0\%\)