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

`m_(SiC)=33.37gSiC`

Step-by-step 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*(1molSiO_2)/(60.08gSiO_2)=0.832molSiO_2\\n_(SiO_2)^(consumed)=50.0gC*(1molC)/(12gC)*(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*(1molSiC)/(1molSiO_2)*(40.11gSiC)/(1molSiC) \\m_(SiC)=33.37gSiC`

Best regards.