Rather than combine 4.00 g of Na₂SO₄ (142.04 g/mol) with a solution containing 10.0 g of H₂SO₄ (98.08 g/mol) ,a student accidentally uses 4.00 g of sodium sulfide (78.05 g/mol) with the acid.The acid reacted with the sodium sulfide to produce highly toxic hydrogen sulfide gas.How many moles of hydrogen sulfide (34.08 g/mol) were produced?
Na₂S(s) + H₂SO₄(aq) $\to$ Na₂SO₄(aq) + H₂S(g)

Question

Rather than combine 4.00 g of Na₂SO₄ (142.04 g/mol) with a solution containing 10.0 g of H₂SO₄ (98.08 g/mol) ,a student accidentally uses 4.00 g of sodium sulfide (78.05 g/mol) with the acid.The acid reacted with the sodium sulfide to produce highly toxic hydrogen sulfide gas.How many moles of hydrogen sulfide (34.08 g/mol) were produced?
Na₂S(s) + H₂SO₄(aq)

\to

Na₂SO₄(aq) + H₂S(g)

Quizplus · Accepted Answer

First, we need to determine the limiting reactant. Using the given masses and molar masses, we find that:

moles of Na2S = 4.00 g / 78.05 g/mol = 0.0512 mol
moles of H2S = 10.0 g / 98.08 g/mol = 0.102 mol

Since Na2S has fewer moles, it is the limiting reactant.

Next, we use the stoichiometry of the balanced equation to find the moles of H2S produced:

1 mol Na2S : 1 mol H2S

Therefore, 0.0512 mol of Na2S reacted to produce 0.0512 mol of H2S.

Finally, we use the molar mass of H2S to convert the moles to grams:

0.0512 mol × 34.08 g/mol = 1.747 g

Therefore, the amount of H2S produced is approximately 1.75 g. The closest answer choice is C, 0.0512 mol.

Rather Than Combine 4 →\to→ Na2SO4(aq)+ H2S(g) A)0

Rather Than Combine 4 $\to$ Na₂SO₄(aq)+ H₂S(g)
A)0