Question

4 moles of sodium carbonate is mixed with 6 moles of $HCl$, which gives sodium chloride by the following reaction.
$N{a}_{2}C{O}_3+2HCl\to 2NaCl+C{O}_2+H_{2}O$
Calculate the amount of excess reagent in moles.

• A. 2 mol
• B. 1 mol
• C. 3 mol
• D. 4 mol

Answer 1

The correct option is B 1 mol

The given balanced chemical reaction is:
$N{a}_{2}C{O}_3+2HCl\to 2NaCl+C{O}_2+H_{2}O$
Here,
Amount of $N{a}_{2}C{O}_3=4$ mol
Amount of $HCl$ = 6 mol
From the balanced equation,
1 mole of $N{a}_{2}C{O}_3$ reacts with 2 moles of $HCl$ to give 2 moles of $NaCl$.
Therefore, 4 moles of sodium carbonate will react with 8 moles of $HCl$.
Thus, $HCl$ is the limiting reagent here.
Since, 2 moles of $HCl$ react with 1 mole of $N{a}_{2}C{O}_3,$
6 moles of $HCl$ will react with 3 moles of $N{a}_{2}C{O}_3$.
Therefore, excess $N{a}_{2}C{O}_3=(4-3)mol=1mol$