How many grams of sulfuric acid are needed to produce 57.0 grams of water? Show all steps of your calculation as well as the final answer.
2NaOH+H2SO4 > 2Na2SO4+2H2O

It is clear that the stichiometry shows that 2.0 moles of NaOH reacts with 1.0 mole of H₂SO₄ to give 2.0 moles of Na₂SO₄ and 2.0 moles of H₂O.

We must convert the grams of water (57.0 g) to moles (n = mass/molar mass).

Now, we can get the number of moles of H₂SO₄ that is needed to produce 3.1666 moles of water.

1.0 mole of H₂SO₄ → 2.0 moles of H₂O, from the stichiometry of the balanced equation.

The number of moles of H₂SO₄ that will produce 3.1666 moles of H₂O (57.0 g) is (1.0 x 3.1666 / 2.0) = 1.5833 moles.

Finally, we should convert the number of moles of H₂SO₄ into grams (n = mass/molar mass).

RomeliaThurston RomeliaThurston · Answer 2 · 2019-01-26T20:14:23+01:00

Answer: 155.134 grams of sulfuric acid is needed.

Explanation:

To calculate the moles, we use the following equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Moles of water:

Given mass of water = 57 grams

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

[tex]\text{Number of moles}=\frac{57g}{18g/mol}=3.166moles[/tex]

For the given chemical reaction, the equation follows:

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

By Stoichiometry of the reaction:

2 moles of water are produced by 1 mole of sulfuric acid

So, 3.166 moles of water will produced by = [tex]\frac{1}{2}\times 3.166=1.583moles[/tex] of sulfuric acid.

Now, to calculate the mass of sulfuric acid, we use the moles equation:

Molar mass of sulfuric acid = 98 g/mol

Putting values in above equation, we get:

[tex]1.583mol=\frac{\text{Given mass}}{98g/mol}[/tex]

Mass of sulfuric acid = 155.134 grams

Hence, 155.134 grams of sulfuric acid is needed.

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