Finding the Limiting Reactant and Calculating Product Yield

How many grams of silicon carbide can be formed from 70.0 g of graphite and 60.0 g of silicon dioxide?

Final answer:

By finding the limiting reactant, silicon dioxide in this case, we can use stoichiometry to determine that 40 grams of silicon carbide can be formed from 70.0 g of graphite and 60.0 g of silicon dioxide.

Explanation:

To find the amount of silicon carbide that can be formed, we first need to find the limiting reactant, the reactant that's completely used up in the reaction and thus determines the maximum amount of product that can be made. In this case, we have graphite (carbon) and silicon dioxide.

First, we need to convert our reactants to moles. The molar mass of carbon (C) is about 12 g/mol, so we have 70.0 g / 12 g/mol = 5.83 mol C. The molar mass of silicon dioxide (SiO2) is about 60 g/mol, and we have 60.0 g / 60 g/mol = 1.00 mol SiO2.

From the balanced chemical equation, SiO2(s) + 2C(s) --> SiC(s) + CO2(g), we see that we need twice as many moles of carbon as silicon dioxide. However, we have more than twice as many moles of carbon than silicon dioxide. Hence, silicon dioxide is the limiting reactant.

To determine the amount of silicon carbide formed, we can use stoichiometry. For every mole of silicon dioxide, one mole of silicon carbide is formed. Hence 1.00 mol SiO2 will yield 1.00 mol SiC. Using the molar mass of SiC (~40 g/mol), this translates to about 40 g.

For further understanding of limiting reactant and stoichiometry concepts, you can visit this link: Limiting Reactant and Stoichiometry.