Question

Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 67.7 g of sulfuric acid is mixed with 83. g of sodium hydroxide. Calculate the minimum mass of sulfuric acid that could be left over by the chemical reaction. Round your answer to significant digits.

Answer 1

Answer: 0.00 grams

Step-by-step explanation:

To calculate the moles, we use the equation:

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

a) moles of sulphuric acid:

`\text{Number of moles}=(67.7g)/(98g/mol)=0.69moles`

b) moles of sodium hydroxide:

`\text{Number of moles}=(83g)/(40g/mol)=2.08moles`

`H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O`

According to stoichiometry :

1 mole of
`H_2SO_4` require 2 moles of
`NaOH`

Thus 0.69 moles of
`H_2SO_4` require=
`(2)/(1)* 0.69=1.38moles` of
`NaOH`

Thus
`H_2SO_4` is the limiting reagent as it limits the formation of product and will be completely used and NaOH is the excess reagent.

Thus sulfuric acid will not be left over by the chemical reaction.