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3 years ago

how many grams of calcium oxide are producer from the decomposition of 125 grams of calcium carbonate

Chemistry

1 answer:

Dmitrij [34]3 years ago

7 0

70.0 g. The decomposition of 125 g CaCO3 produces 700 g CaO.

MM = 100.09 56.08

CaCO3 → CaO + CO2

Mass 125 g

a) Moles of CaCO3 = 125 g CaCO3 x (1 mol CaCO3/100.09 g CaCO3)

= 1.249 mol CaCO3

b) Moles of CaO = 1.249 mol CaCO3 x (1 mol CaO/1 mol CaCO3)

= 1.249 mol CaO

c) Mass of CaO = 1.249 mol CaO x (56.08 g CaO/1 mol CaO) = 70.0 g

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A solution is made by mixing equal masses of methanol, CH4O, and ethanol, C2H6O. Determine the mole fraction of each component t

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Answer: mole fraction of methanol = 0.590

mole fraction of ethanol = 0.410

Explanation:

We are given:

Equal masses of methanol $CH_4O$ and ethanol $C_2H_6O$ are mixed.

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Calculating the moles of methanol in the solution, by using the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{xg}{32.04g/mol}$

Calculating the moles of ethanol in the solution, by using the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{xg}{46.07g/mol}$

To calculate the mole fraction of methanol, we use the equation:

$\chi_{methanol}=\frac{n_{methanol}}{n_{methaol}+n_{ethanol}}$

$\chi_{methanol}=\frac{\frac{xg}{32.04g/mol}}{\frac{xg}{32.04g/mol}+\frac{xg}{46.07g/mol}}=0.590$

To calculate the mole fraction of ethanol, we use the equation:

$\chi_{ethanol}=\frac{n_{ethanol}}{n_{methaol}+n_{ethanol}}$

$\chi_{ethanol}=\frac{\frac{xg}{46.07g/mol}}{\frac{xg}{32.04g/mol}+\frac{xg}{46.07g/mol}}=0.410$

Thus mole fraction of methanol is 0.590 and mole fraction of ethanol 0.410 in three significant figures.

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