When 20 g of CaCO_3 is treated with 20 g of | vedclass

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

Question

(B) The balanced chemical equation is: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$ Molar mass of $CaCO_3 = 100 \ g/mol$ Molar mass of $HCl = 36.5

Vedclass · Accepted Answer

$8.8$

Vedclass · Answer

(B) The balanced chemical equation is: $CaCO_3 + 2HCl ightarrow CaCl_2 + H_2O + CO_2$ Molar mass of $CaCO_3 = 100 \ g/mol$ Molar mass of $HCl = 36.5 \ g/mol$ Molar mass of $CO_2 = 44 \ g/mol$ Moles of $CaCO_3 = \frac{20}{100} = 0.2 \ mol$ Moles of $HCl = \frac{20}{36.5} = 0.548 \ mol$ According to the reaction,$1 \ mol$ of $CaCO_3$ requires $2 \ mol$ of $HCl$. So,$0.2 \ mol$ of $CaCO_3$ requires $0.2 	imes 2 = 0.4 \ mol$ of $HCl$. Since we have $0.548 \ mol$ of $HCl$,$CaCO_3$ is the limiting reagent. Mass of $CO_2$ formed = $0.2 \ mol 	imes 44 \ g/mol = 8.8 \ g$.

When $20 \ g$ of $CaCO_3$ is treated with $20 \ g$ of $HCl$,the mass of $CO_2$ formed would be: (in $g$)

Similar Questions

During the process of fermentation,the number of moles of $CO_2$ liberated from one mole of glucose is

$100 \, g$ $CaCO_3$ reacts with $1 \, L$ $1 \, N$ $HCl$. On completion of reaction,how much weight of $CO_2$ will be obtained? (in $g$)

How many moles of lead chloride will be produced by the reaction of $6.5 \ g$ of $PbO$ and $3.2 \ g$ of $HCl$?

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

The number of moles of $H_2$ required to produce $10$ moles of $NH_3$ in the following reaction is:
$a H_{2(g)} + b NO_{2(g)} \longrightarrow c NH_{3(g)} + d H_2O_{(g)}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module