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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..

Verified Answer :

The balanced reaction that describes the reaction between sulfuric and sodium hydroxide to produce sodium sulfate and water is expressed H2SO4 + 2NaOH → 2H2O + Na2SO4. When 65.5 grams of sodium hydroxide reacts with excess sulfur, we need first to convert the mass to moles and multiply by stoich ratio 2/2 or 1. Hence the moles is multiplied to the molar mass of water. The final answer is 29.475 grams water.

Answer : 29.475 grams of water can be produced.

Solution : Given,

Mass of NaOH = 65.5 g

Molar mass of NaOH = 40 g/mole

Molar mass of \(H_2O\) = 18 g/mole

The given balanced equation is,

\(H_2SO_4+2NaOH\rightarrow 2H_2O+Na_2SO_4\)

First we have to calculate the moles of NaOH.

\(\text{ Moles of NaOH}=\frac{\text{ Mass of NaOH}}{\text{ Molar mass of NaOH}}=\frac{65.5g}{40g/mole}=1.6375moles\)

Now we have to calculate the moles of \(H_2O\).

From the given reaction, we conclude that

As, 2 moles of NaOH react to give 2 moles of \(H_2O\)

So, 1 mole of NaOH react to give 1 mole of \(H_2O\)

And, 1.6375 moles of NaOH react to give 1.6375 moles of \(H_2O\)

The moles of \(H_2O\) = 1.6375 moles

Now we have to calculate the mass of \(H_2O\)

Mass of \(H_2O\) = Moles of \(H_2O\) × Molar mass of \(H_2O\)

Mass of \(H_2O\) = 1.6375 moles × 18 g/mole = 29.475 g

Therefore, 29.475 grams of water can be produced.

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