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3 years ago

How many Mol are in 4.000 grams of MgCl2 ?

Chemistry

1 answer:

grigory [225]3 years ago

4 0

Answer:

The number of mol is: 0, 042 mol in 4 grams of MgCl2

Explanation:

We calculate the weight of 1 mol of MgCl2:

Weight 1mol of MgCl2= weight Mg + (weight Cl)x 2=

24, 3 grams + 2 x 35, 5 grams = 95, 3 grams/mol MgCl2

95, 3 grams------1 mol MgCl2

4 grams -------x = (4 grams x1 mol MgCl2)/ 95, 3 grams= 0, 04197 mol MgCl2

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Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

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According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

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Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

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