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3 years ago

How many grams are contained in a 0.893 mol sample of methane, ch4?

1 answer:

6 0

We are given with a compound, Methane (CH4), with a molar mass of 0.893 mol sample. We are tasked to solve for it's corresponding mass in g. We need to solve first the molecular weight of Methane, that is

C=12 g/mol

H=1g/mol

CH4= 12 g/mol +1(4) g/mol = 16 g/mol

With 0.893 mol sample, its corresponding mass is

g CH4= 0.893 mol x 16g/mol =14.288 g

Therefore, the mass of methane is 14.288 g

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2 years ago

Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of

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Answer:

79%

Explanation:

1) Balanced chemical equation:

2S + 3O₂ → 2SO₃

2) Mole ratio:

2 mol S : 3 mol O₂ : 2 mol SO₃

3) Limiting reactant:

Number of moles of O₂

n = 6.0 g / 32.0 g/mol = 0.1875 mol O₂

Number of moles of S:

n = 7.0 g / 32.065 g/mol = 0.2183 mol S

Ratios:

Actual ratio: 0.1875 mol O₂ / 0.2183 mol S =0.859

Theoretical ratio: 3 mol O₂ / 2 mol S = 1.5

Since there is a smaller proportion of O₂ (0.859) than the theoretical ratio (1.5), O₂ will be used before all S be consumed, and O₂ is the limiting reactant.

4) Calcuate theoretical yield (using the limiting reactant):

0.1875 mol O₂ / x = 3 mol O₂ / 2 mol SO₃

x = 0.1875 × 2 / 3 mol SO₃ = 0.125 mol SO₃

5) Yield in grams:

mass = number of moles × molar mass = 0.125 mol × 80.06 g/mol = 10.0 g

6) Percent yield:

Percent yield, % = (actual yield / theoretical yield) × 100

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3 years ago