Respuesta :
Answer:
13 grams Na₂SO₄ (2 sig.figs.)
Explanation:
1st convert mass values to moles and solve yield using limiting reactant principles and reaction ratio of balance equation.
Determine limiting reactant and
Complete problem by converting yield into grams.
H₂SO₄ + 2NaOH => Na₂SO₄ + 2H₂O
moles H₂SO₄ = 8.83g/98g·mol⁻¹ = 0.0901 mole H₂SO₄
moles NaOH = 9.1g/40g·mol⁻¹ = 0.2275 mole NaOH
Determine Limiting Reactant by dividing each mole value by the respective coefficient in the balanced equation. The smaller value is always the limiting reactant.
H₂SO₄ => (0.0901/1) = 0.0901 <= Limiting Reactant (smaller value)
NaOH => (0.2275/2) = 0.1138
NOTE: when working the problem use the calculated moles of reactant NOT the LR test number. In this problem, use 0.0901 mole H₂SO₄. (yeah, it is the same but this does not occur for the LR in many other problems). Anyways...
H₂SO₄ + 2NaOH => Na₂SO₄ + 2H₂O
moles 0.0901 mole 0.2275 mol 0.0901 mol 2(0.0901 mol)
mass (g) Na₂SO₄ = 0.0901 mole x 142.04 g/mol = 12.798 grams ≅ 13 grams Na₂SO₄ (2 sig.figs.)
Answer:
13 grams Na₂SO₄ (2 sig.figs.)
Explanation:
Firstly
We convert mass values to moles and solve yield using limiting reactant principles and reaction ratio of balance equation.
To determine limiting reactant and
Complete problem by converting yield into grams. We write the equations for the reaction
H₂SO₄ + 2NaOH => Na₂SO₄ + 2H₂O
To give
moles H₂SO₄ = 8.83g/98g·mol⁻¹ = 0.0901 mole H₂SO₄
Then,
moles NaOH = 9.1g/40g·mol⁻¹ = 0.2275 mole NaOH
Determine Limiting Reactant by dividing each mole value by the respective coefficient in the balanced equation. The smaller value is always the limiting reactant.
H₂SO₄ => (0.0901/1) = 0.0901 <= Limiting Reactant (smaller value)
NaOH => (0.2275/2) = 0.1138
N/B: when working the problem use the calculated moles of reactant NOT the LR test number. In this problem, use 0.0901 mole H₂SO₄. (yeah, it is the same but this does not occur for the LR in many other problems). Anyways...
H₂SO₄ + 2NaOH => Na₂SO₄ + 2H₂O
moles 0.0901 mole 0.2275 mol 0.0901 mol 2(0.0901 mol)mass (g)
Na₂SO₄ = 0.0901 mole x 142.04 g/mol = 1
2.798 grams ≅ 13 grams Na₂SO₄ (2 sig.figs.)
