# In the industrial preparation of silicon, 188.5 kg of molten SiO2 reacts with 87.3 kg of carbon to produce 78.4 kg of silicon, according to the following balanced equation what is the percent yield of the reaction? SiO2 (l) + 2C (s) ----> Si (l) + 2CO (g)

Jun 14, 2017

Approx......90%

#### Explanation:

$S i {O}_{2} + C + \Delta \rightarrow S i + 2 C O \left(g\right) \uparrow$

And so we calculate the molar quantities of reactant and product....

"Moles of silica"=(188.5*kgxx10^3*g*kg^-1)/(60.08*g*mol^-1)=3137.5*mol;

"Moles of carbon"=(87.3*kgxx10^3*g*kg^-1)/(12.011*g*mol^-1)=7268.3*mol;

$\text{Moles of silicon} = \frac{78.4 \cdot k g \times {10}^{3} \cdot g \cdot k {g}^{-} 1}{28.1 \cdot g \cdot m o {l}^{-} 1} = 2790.0 \cdot m o l .$

Clearly THERE ARE excess moles of carbon (which makes sense, because it is plentiful and cheap), and thus we calculate yield on the basis of silica as the limiting reagent.........

"% Yield"="Moles of silicon"/"Moles of silica"xx100%

(2790*mol)/(3137.5*mol)xx100%=??%