Question

Aqueous hydrochloric acid will react with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . Suppose 1.5 g of hydrochloric acid is mixed with 2.67 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

Answer 1

Answer: The maximum amount of water that can be produced is 0.74 g

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)

Putting values in equation 1, we get:

`\text{Moles of hydrochloric acid:}=(1.5g)/(36.5g/mol)=0.041mol`

`\text{Moles of sodium hydroxide}=(2.67g)/(40g/mol)=0.067mol`

The chemical equation for the reaction is

`HCl+NaOH\rightarrow NaCl+H_2O`

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of NaOH

So, 0.041 moles of HCl will react with =
`(1)/(1)* 0.041=0.041mol` of NaOH

As, given amount of NaOH is more than the required amount. So, it is considered as an excess reagent. Thus, HCl is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of HCl produces = 1 mole of water

So, 0.041 moles of HCl will produce =
`(1)/(1)* 0.041=0.041moles` of water

Mass of water=
`moles* {\text{Molar Mass}}=0.041mol* 18g/mol=0.74g`

Thus the maximum amount of water that can be produced is 0.74 g