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..

Answer 1

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 2

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}}=(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.