For the given reaction:CaCO_{3} + 2HCl to CaC | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) $1$. Calculate the mass of $HCl$ solution: $	ext{Mass} = 	ext{Density} 	imes 	ext{Volume} = 1.13 \ g \ mL^{-1} 	imes 300 \ mL = 339 \ g$.$2$.

Vedclass · Accepted Answer

$64.97 \ g$ of $HCl$ remains unreacted

Vedclass · Answer

(A) $1$. Calculate the mass of $HCl$ solution: $	ext{Mass} = 	ext{Density} 	imes 	ext{Volume} = 1.13 \ g \ mL^{-1} 	imes 300 \ mL = 339 \ g$.$2$. Calculate the mass of pure $HCl$: $	ext{Mass of } HCl = 339 \ g 	imes 0.3855 = 130.68 \ g$.$3$. Calculate moles of reactants:Moles of $CaCO_{3} = \frac{90 \ g}{100 \ g \ mol^{-1}} = 0.90 \ mol$.Moles of $HCl = \frac{130.68 \ g}{36.5 \ g \ mol^{-1}} \approx 3.58 \ mol$.$4$. Determine the Limiting Reagent $(LR)$:According to the stoichiometry, $1 \ mol$ of $CaCO_{3}$ reacts with $2 \ mol$ of $HCl$.For $0.90 \ mol$ of $CaCO_{3}$, we need $0.90 	imes 2 = 1.80 \ mol$ of $HCl$.Since we have $3.58 \ mol$ of $HCl$, $CaCO_{3}$ is the limiting reagent.$5$. Calculate remaining $HCl$:$HCl$ consumed $= 1.80 \ mol$.$HCl$ remaining $= 3.58 - 1.80 = 1.78 \ mol$.Mass of $HCl$ remaining $= 1.78 \ mol 	imes 36.5 \ g \ mol^{-1} = 64.97 \ g$.

Similar Questions

One mole of nitrogen gas on reaction with $3.01 \times 10^{23}$ molecules of hydrogen gas produces

$KMnO_4$ reacts with $KI$ in basic medium to form $I_2$ and $MnO_2$. When $250 \ mL$ of $0.1 \ M \ KI$ solution is mixed with $250 \ mL$ of $0.02 \ M \ KMnO_4$ in basic medium, what is the number of moles of $I_2$ formed?

For the reaction $A + 2B \rightarrow C$,$5 \ \text{mol}$ of $A$ and $8 \ \text{mol}$ of $B$ produce how many moles of $C$?

$50.0 \ kg$ of $N_{2(g)}$ and $10.0 \ kg$ of $H_{2(g)}$ are mixed to produce $NH_{3(g)}$. Calculate the amount of $NH_{3(g)}$ formed. Identify the limiting reagent in the production of $NH_{3(g)}$ in this situation.

Vedclass Test Series

Exam Paper Generator

Online Exam Module