Question

If 49 grams of H2SO4 react with 80 grams of NaOH, how much reactant will be leftover after the reaction is complete?
A) 24.5g of H2SO4
B) none of either compound
C) 20 g of NaOH
D) 40 g of NaOH
E) 60 g of NaOH

Answer 1

Answer on Question #45025 – Chemistry – Other

Question

If 49 grams of $\mathrm{H}_2\mathrm{SO}_4$ react with 80 grams of NaOH, how much reactant will be leftover after the reaction is complete?

A) $24.5\mathrm{g}$ of $\mathrm{H}_2\mathrm{SO}_4$

B) none of either compound

C) $20\mathrm{g}$ of NaOH

D) $40\mathrm{g}$ of NaOH

E) $60\mathrm{g}$ of NaOH

Solution

The chemical equation is as follows

$$\mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + 2\mathrm{H}_2\mathrm{O}$$

Molar masses of the reactants are M(NaOH) = 40 g/mol, M(H₂SO₄) = 98 g/mol.

As is clear from the reaction stoichiometry 1 mol of $\mathrm{H}_2\mathrm{SO}_4$ react with 2 mol of NaOH, or 1 mol·98 g/mol = 98 g $\mathrm{H}_2\mathrm{SO}_4$ react with 2 mol·40 g/mol = 80 g NaOH, so we can write the following proportion:

98 g $(\mathrm{H}_2\mathrm{SO}_4) - 80\mathrm{g}$ (NaOH)

49 g $(\mathrm{H}_2\mathrm{SO}_4) - X\mathrm{g}$ (NaOH), whence

$X = 49 \cdot 80 / 98 = 40\mathrm{g}$ – mass of NaOH needed to completely neutralize $49\mathrm{g}$ of $\mathrm{H}_2\mathrm{SO}_4$.

Since $80\mathrm{g}$ of NaOH is taken, $80 - 40 = 40\mathrm{g}$ of NaOH will be leftover after the reaction is complete.

Answer: D) $40\mathrm{g}$ of NaOH

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