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How many grams of sulfuric acid are needed to react completely with 39.9 g of sodium hydroxide?

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Final answer:

To completely react with 39.9 g of NaOH, you would need approximately 195.7 g of sulfuric acid.

Step-by-step explanation:

In order to determine the grams of sulfuric acid needed to react completely with sodium hydroxide, we need to balance the chemical equation:

H2SO4 + 2NaOH → Na2SO4 + 2H2O

From the equation, we can see that 1 mole of H2SO4 reacts with 2 moles of NaOH. First, we need to determine the number of moles of NaOH:

Moles of NaOH = mass of NaOH / molar mass of NaOH

Molar mass of NaOH = 23.0 g/mol (Na) + 16.0 g/mol (O) + 1.0 g/mol (H) = 40.0 g/mol

Moles of NaOH = 39.9 g / 40.0 g/mol = 0.9975 mol

Since the molar ratio between H2SO4 and NaOH is 1:2, the number of moles of H2SO4 needed is twice the number of moles of NaOH:

Moles of H2SO4 = 2 * 0.9975 mol = 1.995 mol

Finally, we can calculate the grams of H2SO4:

Grams of H2SO4 = moles of H2SO4 * molar mass of H2SO4

Molar mass of H2SO4 = 2.0 g/mol (H) + 32.1 g/mol (S) + 16.0 g/mol (O) * 4 = 98.1 g/mol

Grams of H2SO4 = 1.995 mol * 98.1 g/mol = 195.7 g

User Shamal Perera
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