Question

Calculate the mas of Magnesium required to completely react with 250 cm3 of 0.1 M HCl.
GIVEN : Mg + 2HCl --> MgCl2 + H2

Answer 1

Explanation:

Given,

$Mg + 2HCl \rightleftharpoons \: MgCl_2 + H_2$

To Find :-

Mass of Magnesium required to completely react with 250 cm^3 of 0.1M HCl.

Solution :-

By seeing the Chemical reaction we can say that :-

24g of Mg reacts with 2HCl

$\sf\implies$ 24g of Mg reacts with 2 moles of HCl. , Generally we take this quantity as per liter.

[Since, 1 litre = 1000milli litres.]

But in the question they asked us to find the amoung of Mg reacted with 0.1M of HCl

$\sf\implies$

$= \dfrac{24g \times 0.1 \: M } {2 M}$

$= 12g \times 0.1$

$= 1.2g$

Since , amount of Mg reacted with 0.1M of HCl = 1.2g

But , this 1.2 g of Mg reacted with 1 litre(1000cm^3)

In question they asked to find amount of Mg reacted with 250cm^3

$= \dfrac{250 {cm}^{3} \times 1.2g }{1000 {cm}^{3} }$

$= \dfrac{1.2g}{4}$

= 0.3g

Since, mass of Magnesium required to completely react with 250 cm^3 of 0.1 M HCl = 0.3g