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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Answer: (i) 0.261 moles of HCl, (ii) 0.131 moles of [tex]MgCl_2[/tex] and (iii) 12.5 grams of [tex]MgCl_2[/tex]
Explanation: The balanced equation for the formation of HCl by the reaction magnesium chloride with water is:
[tex]MgCl_2+2H_2O\rightarrow 2HCl+Mg(OH)_2[/tex]
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.
[tex]5.85LHCl(\frac{1mol}{22.4L})[/tex]
= 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:
[tex]0.261molHCl(\frac{1molMgCl_2}{2molHCl})[/tex]
= 0.131 mol [tex]MgCl_2[/tex]
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.
[tex]0.131molMgCl_2(\frac{95.2g}{1mol})[/tex]
= 12.5 g [tex]MgCl_2[/tex]
