Sulfur dioxide gas (so2) reacts with excess oxygen gas (o2) and excess liquid water (h2o) to form liquid sulfuric acid (h2so4). in the laboratory, a chemist carries out this reaction with 67.2 l of sulfur dioxide and gets 250 g of sulfuric acid. • calculate the theoretical yield of sulfuric acid. • calculate the percent yield of the reaction. (one mole of any gas occupies 22.4 l under certain conditions of temperature and pressure. assume those conditions for this question.)

1) Answer is: the percent yield of the reaction is 84.5%.

Balanced chemical reaction: 2SO₂ + 2H₂O + O₂ → 2H₂SO₄.

V(SO₂) = 67.2 L; voulume of sulfur dioxide.

Vm = 22.4 L/mol; molar volume.

m(H₂SO₄) = 250 g; actual mass of sulfuric acid.

n(SO₂) = V(SO₂) ÷ Vm.

n(SO₂) = 67.2 L ÷ 22.4 L/mol.

n(SO₂) = 3 mol; amount of sulfur dioxide.

2) From balanced chemical reaction: n(SO₂) : n(H₂SO₄) = 2 : 2 (1 : 1).

n(H₂SO₄) = n(SO₂).

n(H₂SO₄) = 3 mol; amoun of sulfuric acid.

M(H₂SO₄) = 98.08 g/mol; molar mass of sulfuric acid.

m(H₂SO₄) = n(H₂SO₄) · M(H₂SO₄).

m(H₂SO₄) = 3 mol · 98.08 g/mol.

m(H₂SO₄) = 294.24 g; theoretical mass of sulfuric acid.

Theoretical yield is the maximum amount of product that can be produced from limiting reactant and actual yield is a product that is obtained by experimentation.

the percent yield = actual yield / theoretical yield.

the percent yield = 250 g ÷ 294.24 g · 100%.

the percent yield = 84.5 %.

answer: mercury

explanation:

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )