Calculate the mass of MgCl2 in the following reaction of Mg and HCl, when 17.8g of Mg is added?

First, we need to write the balanced chemical equation for the reaction between Mg and HCl:

Mg + 2HCl -> MgCl2 + H2

From the balanced equation, we can see that 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of MgCl2.

1 mole of Mg has a mass of 24.3g, so 17.8g of Mg is equal to 17.8g / 24.3g/mol = 0.733 moles of Mg.

Since the mole ratio between Mg and MgCl2 is 1:1, the number of moles of MgCl2 produced will also be 0.733 moles.

The molar mass of MgCl2 is 95.21g/mol, so the mass of MgCl2 produced will be:

0.733 moles * 95.21g/mol = 69.8g

Therefore, the mass of MgCl2 produced when 17.8g of Mg is added is 69.8g.

The balanced chemical reaction is:

```
Mg + 2 HCl -> MgCl2 + H2
```

From the balanced equation, we can see that 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of MgCl2.

The molar mass of Mg is 24.31 g/mol, and the molar mass of MgCl2 is 95.21 g/mol.

Therefore, 17.8 g of Mg is equal to:

```
17.8 g Mg / 24.31 g/mol = 0.732 mol Mg
```

According to the stoichiometry of the balanced equation, 0.732 mol of Mg will react with 1.464 mol of HCl to produce 0.732 mol of MgCl2.

Therefore, the mass of MgCl2 produced is:

```
0.732 mol MgCl2 - 95.21 g/mol = 69.7 g MgCl2
```

Therefore, the mass of MgCl2 in the reaction is 69.7 g.