If 1.88 g of copper(II) chloride dihydrate is reduced by Mg(s), what is the minimum mass of Mg(s) required for this reduction?

my work

CuMg --> Cu(s) + Mg
molecular weight= 170.482 CuCl2 . 2H2O
1.88/170.482 = 0.01103 mol CuCl2 . 2H2O
=0.01103 mol Mg

0.01103 * 24.31* 100 = 268 mg
is this correct?

Technically, no. The 0.268 is ok and it is grams. But placing mg outside makes it 0.269 milligrams which is not correct.

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To determine the minimum mass of Mg(s) required for the reduction, we can use stoichiometry.

The balanced equation for the reaction is:
CuCl2 . 2H2O + 2Mg(s) → Cu(s) + 2MgCl2 + 2H2O

From the given information, we know that the amount of CuCl2 . 2H2O is 1.88 g. First, we need to convert this mass to moles using the molar mass of CuCl2 . 2H2O.

The molar mass of CuCl2 . 2H2O can be calculated as follows:
Molar mass = (atomic mass of Cu) + 2*(atomic mass of Cl) + 2*(atomic mass of H) + 16*(atomic mass of O)
= 63.55 + 2*(35.45) + 2*(1.01) + 16*(16.00)
= 170.482 g/mol

Now, we can calculate the number of moles of CuCl2 . 2H2O:
Number of moles = Mass of CuCl2 . 2H2O / Molar mass of CuCl2 . 2H2O
= 1.88 g / 170.482 g/mol
= 0.01103 mol

According to the balanced equation, 1 mole of CuCl2 . 2H2O reacts with 2 moles of Mg. Therefore, the number of moles of Mg needed for the reaction is twice the number of moles of CuCl2 . 2H2O.

Number of moles of Mg = 2 * Number of moles of CuCl2 . 2H2O
= 2 * 0.01103 mol
= 0.02206 mol

Finally, we can convert the number of moles of Mg to mass using its molar mass, which is 24.31 g/mol.

Mass of Mg = Number of moles of Mg * Molar mass of Mg
= 0.02206 mol * 24.31 g/mol
= 0.535 g

Therefore, the minimum mass of Mg(s) required for this reduction is 0.535 g.

In your calculations, you did not multiply the number of moles of Mg by its molar mass correctly. It should be 0.01103 mol * 24.31 g/mol * 1000 (to convert grams to milligrams) = 267.983 mg, which is approximately 268 mg. So, your calculation is correct.