Question

A piece of Mg is dissolved in 40 ml of 0.1N HCl completely. The excess of acid was neutralized by 15 ml of 0.2N NaOH.The weight of Mg is.

Answer 1

me of HCl = 40 x 0.1 = 4 Excess me of acid = 15 x 0.2 = 3 Therefore me of acid that dissolved Mg = me of Mg dissolved = 4-3 = 1 me = .001 eq of Mg = 0.012 g of Mg

Answer 2

The weight of magnesium is 0.012 gm.  

Explanation:

From the given,

For HCl:

Volume = 40 ml

Conversion of ml into litres

$\frac{40}{1000}=0.040 \mathrm{L}$

N = 0.1 M

Moles of HCL=$0.040 \times 0.1\ \mathrm{M}=0.0040\ \mathrm{m}$

For NaOH:

Volume = 15 ml

Conversion ml into litres = $\frac{15}{1000}=0.015\ \mathrm{L}$

N = 0.2 M

Moles of NaOH=0.015 \mathrm{L}×0.2 \mathrm{M}=0.0030 \mathrm{m}

Excess moles of HCl = 0.0040 - 0.0030 = 0.0010 moles

Balanced equation between Magnesium and HCl is as follows,

$\mathrm{Mg}+2 \mathrm{HCl} \rightarrow \mathrm{MgCl}_{2}+\mathrm{H}_{2}$

Therefore,the ratio between Mg and HCl is 1:2

Moles of Mg=$\frac{\text { Excess of }\mathrm{HCl}}{2}=\frac{0.0010}{2}=0.0005\ \mathrm{m}$

Molar mass of Mg=24.305 g/mol

Mass of Mg=$0.0005 \times 24.305\ \frac{\mathrm{g}}{\mathrm{mol}}=0.012\ \mathrm{g}$