Question

For the following reaction, 11.6 grams of sulfur are allowed to react with 23.5 grams of carbon monoxide . sulfur(s) + carbon monoxide(g) sulfur dioxide(g) + carbon(s) What is the maximum amount of sulfur dioxide that can be formed? grams What is the FORMULA for the limiting reagent? CO What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

Answer: The formula of limiting reagent is 'S', the amount of excess reagent (carbon monoxide) left is 3.22 grams and the maximum amount of sulfur dioxide formed is 23.2 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

• For sulfur:

Given mass of sulfur = 11.6 g

Molar mass of sulfur = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of sulfur}=\frac{11.6g}{32g/mol}=0.3625mol$

• For carbon monoxide:

Given mass of carbon monoxide = 23.5 g

Molar mass of carbon monoxide = 28 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon monoxide}=\frac{23.5g}{28g/mol}=0.840mol$

The chemical equation for the reaction of sulfur and carbon monoxide follows:

$S(s)+2CO(g)\rightarrow SO_2(g)+2C(s)$

By Stoichiometry of the reaction:

1 mole of sulfur reacts with 2 moles of carbon monoxide

So, 0.3625 moles of sulfur will react with = $\frac{2}{1}\times 0.3625=0.725mol$ of carbon monoxide

As, given amount of carbon monoxide is more than the required amount. So, it is considered as an excess reagent.

Thus, sulfur is considered as a limiting reagent because it limits the formation of product.

Amount of excess reagent (carbon monoxide) = [0.840 - 0.725] = 0.115 moles

By Stoichiometry of the reaction:

1 mole of sulfur produces 1 mole of sulfur dioxide

So, 0.3625 moles of sulfur will produce = $\frac{1}{1}\times 0.3625=0.3625moles$ of sulfur dioxide

Now, calculating the mass of carbon monoxide and sulfur dioxide from equation 1, we get:

• For carbon monoxide:

Molar mass of carbon monoxide = 28 g/mol

Excess moles of carbon monoxide = 0.115 moles

Putting values in equation 1, we get:

$0.115mol=\frac{\text{Mass of carbon monoxide}}{28g/mol}\\\\\text{Mass of carbon monoxide}=(0.115mol\times 28g/mol)=3.22g$

• For sulfur dioxide:

Molar mass of sulfur dioxide = 64 g/mol

Excess moles of sulfur dioxide = 0.3625 moles

Putting values in equation 1, we get:

$0.3625mol=\frac{\text{Mass of sulfur dioxide}}{64g/mol}\\\\\text{Mass of sulfur dioxide}=(0.3625mol\times 64g/mol)=23.2g$

Hence, the formula of limiting reagent is 'S', the amount of excess reagent (carbon monoxide) left is 3.22 grams and the maximum amount of sulfur dioxide formed is 23.2 grams