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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..

Respuesta :

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 [tex]H_2O[/tex] = 18 g/mole

The given balanced equation is,

[tex]H_2SO_4+2NaOH\rightarrow 2H_2O+Na_2SO_4[/tex]

First we have to calculate the moles of NaOH.

[tex]\text{ Moles of NaOH}=\frac{\text{ Mass of NaOH}}{\text{ Molar mass of NaOH}}=\frac{65.5g}{40g/mole}=1.6375moles[/tex]

Now we have to calculate the moles of [tex]H_2O[/tex].

From the given reaction, we conclude that

As, 2 moles of NaOH react to give 2 moles of [tex]H_2O[/tex]

So, 1 mole of NaOH react to give 1 mole of [tex]H_2O[/tex]

And, 1.6375 moles of NaOH react to give 1.6375 moles of [tex]H_2O[/tex]

The moles of [tex]H_2O[/tex] = 1.6375 moles

Now we have to calculate the mass of [tex]H_2O[/tex]

Mass of [tex]H_2O[/tex] = Moles of [tex]H_2O[/tex] × Molar mass of [tex]H_2O[/tex]

Mass of [tex]H_2O[/tex] = 1.6375 moles × 18 g/mole = 29.475 g

Therefore, 29.475 grams of water can be produced.