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2 years ago

How many liters of a 0.352 M solution of Ca(SO4) would contain 62.1 g of Ca(SO4)?

Chemistry

1 answer:

konstantin123 [22]2 years ago

4 0

Answer:

1.3 L.

Explanation:

Molarity is the no. of moles of solute per 1.0 L of the solution.

M = (no. of moles of CaSO₄)/(Volume of the solution (L))

M = 0.352 M.

no. of moles of CaSO₄ = mass/molar mass = (62.1 g / 136.14 g/mol) = 0.456 mol,

Volume of the solution = ??? L.

∴ (0.352 M) = (0.456 mol)/(Volume of the solution (L))

∴ (Volume of the solution (L) = (0.456 mol)/(0.352 M) = 1.296 L ≅ 1.3 L.

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Answer:

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2. 217.4 g O₂

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Explanation:

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120 g SiCl₄ x 1 mol/169.9 g = .706 mol SiCl₄

Moles to moles:

For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

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Theoretical yield:

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Leftover reactant for trial 1

We know oxygen is the excess reactant.

Mass to moles:

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240 g O₂ x 1 mol/32.00 g = 7.5 mol O₂

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Moles to mass:

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Mass to moles:

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75 g SiCl₄ x 1 mol/169.9 g = .441 mol SiCl₄

Moles to moles:

For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

.441 mol SiCl₄ = .441 mol SiO₂

Moles to mass:

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Theoretical yield:

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How many liters of a 0.352 M solution of Ca(SO4) would contain 62.1 g of Ca(SO4)?​

How many liters of a 0.352 M solution of Ca(SO4) would contain 62.1 g of Ca(SO4)?