Question

How many grams of water can be produced when 65.5 grams of sodium hydroxide reacts with excess sulfuric acid?

Unbalanced equation: H2SO4 + NaOH → H2O + Na2SO4

Show, or explain, all of your work along with the final answer.

Accepted Answer

How many grams of water can be produced when 65.5 grams of sodium hydroxide reacts with excess sulfuric acid?

Unbalanced equation: $\mathrm{H}_2\mathrm{SO}_4 + \mathrm{NaOH} \rightarrow \mathrm{H}_2\mathrm{O} + \mathrm{Na}_2\mathrm{SO}_4$

Show, or explain, all of your work along with the final answer.

Solution:

\mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{NaOH} \rightarrow 2\mathrm{H}_2\mathrm{O} + \mathrm{Na}_2\mathrm{SO}_4

Find the amount of the substance sodium hydroxide:

n(\mathrm{NaOH}) = \frac{m(\mathrm{NaOH})}{M(\mathrm{NaOH})}

n(\mathrm{NaOH}) = \frac{65.5g}{40g/\mathrm{mole}} = 1.64\ \mathrm{mole}

Find the equation of the reaction mass of water which formed. Since sulfuric acid is reacted in excess, we find that the mass of water by weight sodium hydroxide:

\frac{n(\mathrm{NaOH})}{n(\mathrm{H}_2\mathrm{O})} = \frac{2}{2} = \frac{1}{1}

\frac{m(\mathrm{NaOH})}{M(\mathrm{NaOH})} = \frac{m(\mathrm{H}_2\mathrm{O})}{M(\mathrm{H}_2\mathrm{O})}

m(\mathrm{H}_2\mathrm{O}) = \frac{m(\mathrm{NaOH}) \cdot M(\mathrm{H}_2\mathrm{O})}{M(\mathrm{NaOH})}

m(\mathrm{H}_2\mathrm{O}) = \frac{65.5g \cdot 18g/\mathrm{mole}}{40g/\mathrm{mole}} = 29.475g

Answer: 29.475 g of water.

Question #30925

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