Answer:
2.16 g H₂O
Step-by-step explanation:
Since we are not given the limiting reagent, we need to convert both reactants into the product. The actual amount of product will be the smaller mass produced.
To find the answer, we need to (1) convert grams reactant to moles reactant (via molar mass), then (2) convert moles reactant to moles H₂O (via mole-to-mole ratio from balanced equation coefficients), and then (3) convert moles H₂O to grams H₂O (via molar mass). It is important to arrange the ratios in a way that allows for the cancellation of units.
The balanced equation:
1 H₂SO₄(aq) + 2 NaOH(s) -----> 1 Na₂SO₄(aq) + 2 H₂O(l)
Molar Mass (H₂SO₄): 2(1.008 g/mol) + 32.065 g/mol + 4(15.998 g/mol)
Molar Mass (H₂SO₄): 98.073 g/mol
Molar Mass (NaOH): 22.990 g/mol + 15.998 g/mol + 1.008 g/mol
Molar Mass (NaOH): 39.996 g/mol
Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol
Molar Mass (H₂O): 18.014 g/mol
5.88 g H₂SO₄ 1 mole 2 moles H₂O 18.014 g
---------------------- x ----------------- x ------------------------ x ----------------- =
98.073 g 1 mole H₂SO₄ 1 mole
= 2.16 g H₂O
6.3 g NaOH 1 mole 2 moles H₂O 18.014 g
-------------------- x ----------------- x -------------------------- x ---------------- =
39.996 g 2 moles NaOH 1 mole
= 2.84 g H₂O
Since H₂SO₄ produces the smallest amount of product, it is the limiting reagent. In this case, H₂SO₄ is completely used up before NaOH has the chance to react totally. Therefore, the actual amount of H₂O produced is 2.16 g.