Question

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

Answer 1

Answer:
`O_2`

Step-by-step explanation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Moles of
`CS_2=(100g)/(76g/mol)=1.3moles`

Moles of
`O_2=(38g)/(32g/mol)=1.2moles`

For the given chemical reaction, the equation follows:

`CS_2+O_2\rightarrow CO_2+2S`

By Stoichiometry:

1 mole of oxygen reacts with 1 mole of carbon disulphide

So, 1.2 moles of oxygen reacts with =
`(1)/(1)* 1.2=1.2moles` 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.

Answer 2

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.