Question

When magnesium chloride reacts with water, 5.85 L HCl(g) is produced.

Answer 1

Answer: (i) 0.261 moles of HCl, (ii) 0.131 moles of $MgCl_2$ and (iii) 12.5 grams of $MgCl_2$

Explanation: The balanced equation for the formation of HCl by the reaction magnesium chloride with water is:

$MgCl_2+2H_2O\rightarrow 2HCl+Mg(OH)_2$

From above balanced equation, there is 1:2 mol ratio between magnesium chloride and HCl.

As per the given information, 5.85 L of HCl gas is produced. here, temperature and pressure is not mentioned, so let's assume STP conditions.

At STP, volume of 1 mol of a gas is 22.4 L. With the help of this, we can calculate the moles of HCl present in 5.85 L.

$5.85LHCl(\frac{1mol}{22.4L})$

= 0.261 mol HCl

As there is 1:2 mol ratio between magnesium chloride and HCl, we can calculate the moles of magnesium chloride from the above calculated moles of HCl as:

$0.261molHCl(\frac{1molMgCl_2}{2molHCl})$

= 0.131 mol $MgCl_2$

Molar mass of magnesium chloride is given as 95.2 gram per mol. On multiplying the moles by molar mass we can calculate its mass reacted.

$0.131molMgCl_2(\frac{95.2g}{1mol})$

= 12.5 g $MgCl_2$

Answer 2

Answer: How many moles of HCl was produced?

⇒ 0.261 moles of HCl

How many moles of MgCl2 reacted?

⇒ 0.131 moles of MgCl2

What mass of MgCl2 reacted?

⇒ 12.4 g MgCl2

Explanation:

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