Question

3. Silicon carbide, SiC, is prepared by heating silicon dioxide in the presence of graphite. Carbon dioxide is the by-product of the reaction. How many grams of silicon carbide can be formed from the reaction of 50.0 grams of graphite with 50.0 grams of silicon dioxide

Answer 1

Answer:

$m_{SiC}=33.37gSiC$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$SiO_2+2C\rightarrow SiC+CO_2$

Thus, by stoichiometry, the limiting reagent is computed by comparing the available moles of silicon dioxide and the consumed moles of silicon dioxide by graphite as shown below:

$n_{SiO_2}^{available}=50.0gSiO_2*\frac{1molSiO_2}{60.08gSiO_2}=0.832molSiO_2\\n_{SiO_2}^{consumed}=50.0gC*\frac{1molC}{12gC}*\frac{1molSiO}{2molC}=2.08molSiO_2$

In such a way, since there will be less available silicon dioxide, it is the limiting reagent, therefore, the grams of silicon carbide, turns out:

$m_{SiC}=0.832molSiO_2*\frac{1molSiC}{1molSiO_2}*\frac{40.11gSiC}{1molSiC} \\m_{SiC}=33.37gSiC$

Best regards.