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Carbon disulfide (CS2) undergoes a single displacement reaction with O2 to form CO2. If 100 grams of CS2 reacts with 38 grams of O2, what will the limiting reagent be?
CS2 + O2 CO2 + 2S

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The balanced chemical reaction is:

CS2 + O2 = CO2 + 2S

We are given the amounts of the reactants. These amounts will be the starting point for the calculations. First, we convert these amounts into moles.

100 g CS2 (1 mol / 76.15 g ) = 1.3132 mol CS2
38 g O2 (1 mol / 32 g) = 1.1875 mol O2

From the reaction, the mole ratio of the reactants is 1:1. Therefore, the limiting reactant is CS2.

Answer: [tex]O_2[/tex]

Explanation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Moles of [tex]CS_2=\frac{100g}{76g/mol}=1.3moles[/tex]

Moles of [tex]O_2=\frac{38g}{32g/mol}=1.2moles[/tex]

For the given chemical reaction, the equation follows:

[tex]CS_2+O_2\rightarrow CO_2+2S[/tex]

By Stoichiometry:

1 mole of  oxygen reacts with 1 mole of carbon disulphide

So, 1.2 moles of oxygen reacts with = [tex]\frac{1}{1}\times 1.2=1.2moles[/tex] of carbon disulphide

As, the required amount of carbon disulphide is more than the required amount. Hence, it is considered as the excess reagent.  (1.3-1.2)= 0.1 mole will be left unused.

Oxygen is considered as a limiting reagent because it limits the formation of product.

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