Answer:
[tex]m_{SiC}=33.37gSiC[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]SiO_2+2C\rightarrow SiC+CO_2[/tex]
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:
[tex]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[/tex]
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:
[tex]m_{SiC}=0.832molSiO_2*\frac{1molSiC}{1molSiO_2}*\frac{40.11gSiC}{1molSiC} \\m_{SiC}=33.37gSiC[/tex]
Best regards.
