Question

Consider the following reaction: 2Na 2HCI > 2N»C1 + H2 How many mols of hydrogen gas (H2) can be produced if you begin with 13.08 grams of each reactant?

Answer 1

Answer: The moles of hydrogen gas that can be formed are 0.18 moles.

Step-by-step explanation:

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

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

• For Sodium metal:

Given mass of sodium metal = 13.08 g

Molar mass of sodium metal = 23 g/mol

Putting values in above equation, we get:

`\text{Moles of sodium metal}=(13.08g)/(23g/mol)=0.57mol`

• For hydrochloric acid:

Given mass of hydrochloric acid = 13.08 g

Molar mass of hydrochloric acid = 36.5 g/mol

Putting values in above equation, we get:

`\text{Moles of hydrochloric acid}=(13.08g)/(36.5g/mol)=0.36mol`

For the given chemical equation:

`2Na+2HCI\rightarrow 2NaCl+H_2`

By Stoichiometry of the reaction:

2 moles of hydrochloric acid reacts with 2 moles of sodium metal.

So, 0.36 moles of hydrochloric acid will react with =
`(2)/(2)* 0.36=0.36moles` of sodium metal.

As, given amount of sodium metal is more than the required amount. Thus, it is considered as an excess reagent.

So, hydrochloric acid is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the above reaction:

2 moles of hydrochloric acid is producing 1 moles of hydrogen gas.

So, 0.36 moles of hydrochloric acid will produce =
`(1)/(2)* 0.36=0.18moles` of hydrogen gas.

Hence, the moles of hydrogen gas that can be formed are 0.18 moles.