22.0cm3 of a solution containing 0.1mol of HCl per dm3 of a solution neutralizes 20.0cm3 of a solution 7.6g of anhydrous base M2C03 per dm3 of solution. (a) write a balance equation for the reaction calculate; (1) the molar conc. of the base and (11) molar mass of the base?

Faheemadeiza2012

2 years ago

Sure, I can help with this.

**Balanced equation:**

```
HCl + M2CO3 → M2Cl2 + CO2 + H2O
```

**Calculations:**

(1) Molar concentration of the base:

```
n(HCl) = c(HCl)V(HCl) = 0.1 mol/dm³ × 22.0 cm³ × 1 dm³/1000 cm³ = 0.0022 mol
```

```
n(M2CO3) = n(HCl) = 0.0022 mol
```

```
c(M2CO3) = n(M2CO3)/V(M2CO3) = 0.0022 mol / 20.0 cm³ × 1 dm³/1000 cm³ = 0.11 mol/dm³
```

(2) Molar mass of the base:

```
M(M2CO3) = m(M2CO3)/n(M2CO3) = 7.6 g / 0.0022 mol = 3454.55 g/mol
```

Therefore, the molar concentration of the base is 0.11 mol/dm³ and the molar mass of the base is 3454.55 g/mol.