Question

How many moles of excess reagent are left by the reaction of 6.4 mol of $HCl$ and 3.5 mol of Zn?
The reaction that takes place is:
$Zn+2HCl\to ZnC{l}_2+H_2$

• A. 0.30 mol
• B. 3.45 mol
• C. 4.90 mol
• D. 0.54 mol

Answer 1

The correct option is A 0.30 mol

$Zn+2HCl\to ZnC{l}_2+H_2$
To find limiting reagent:
For $Zn=\frac{3.5}{1}=3.5$
For $HCl=\frac{6.4}{2}=3.2$
From above, we can say that HCl is the limiting reagent and Zn is the excess reagent.
To find the number of moles of Zn required to consume 6.4 mol HCl:

$\frac{6.4molofHCl\times 1molofZnC{l}_2}{2molofHCl}=3.2molofZn$
Number of moles of the excess reagent left = 3.5 - 3.2 = 0.3 mol