Hello! My question below is silimar to one expected on our upcoming exam. If you

Question

Hello! My question below is silimar to one expected on our upcoming exam. If you would be willing to explain the process of how to answer this kind of problem, it would be extreamly helpful. Dumb it down as much as possible, please! Thanks so much!

The abrasive silicon carbide, SiC (MM 40.1g/mol), is made by the reaction of silicon dioxide, SiO2 (MM 60.1g/mol), with graphite, C (MM 12.0g/mol).

Note: 1SiO2(s)+3C(s)--->1SiC(s)+2CO(g)

150.0g silicon dioxide is mixed with 101.5g graphite. Determine the limiting reactant, and the theoretical yield of silicon carbide.

Explanation / Answer

From the reaction you can observe that 1 mole of SiO2 reacts with 3 moles of Carbon graphite .

The ratio of carbon to SiO2 = 3/1 = 3

Now first calculate moles of SiO2 = 150 g / 60.1 g/mol. == 2.495 mol

Moles of carbon graphite = 101.5 g /12 g/mol = 8.458 mol

The ratio of carbon to SiO2 = 8.458/2.495 =3.4

The above ratio is greater than 3

This means we have excess of carbon than SiO2 .

So. The limiting reactant will be SiO2

Yield of SiC =

For 1 mole of SiO2 reacted to get 1 mol of SiC

Here 2.495 moles reacted to get 2.495miles of SiC

Yield of SiC = 2.495 mol x 40.1 g/mol = 100.0 g

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