What will be the weight of lime (CaO) obtaine | vedclass

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

Question

(D) The chemical reaction for the decomposition of limestone is: $CaCO_3(s) \longrightarrow CaO(s) + CO_2(g)$Mass of pure $CaCO_3 = 300 \ kg 	imes 0.

Vedclass · Accepted Answer

$151.2$

Vedclass · Answer

(D) The chemical reaction for the decomposition of limestone is: $CaCO_3(s) \longrightarrow CaO(s) + CO_2(g)$Mass of pure $CaCO_3 = 300 \ kg 	imes 0.90 = 270 \ kg = 270,000 \ g$.Molar mass of $CaCO_3 = 40 + 12 + (3 	imes 16) = 100 \ g/mol$.Moles of $CaCO_3 = \frac{270,000 \ g}{100 \ g/mol} = 2700 \ mol$.According to the stoichiometry,$1 \ mol$ of $CaCO_3$ produces $1 \ mol$ of $CaO$.Therefore,$2700 \ mol$ of $CaCO_3$ produces $2700 \ mol$ of $CaO$.Molar mass of $CaO = 40 + 16 = 56 \ g/mol$.Weight of $CaO = 2700 \ mol 	imes 56 \ g/mol = 151,200 \ g = 151.2 \ kg$.

What will be the weight of lime $(CaO)$ obtained by heating $300 \ kg$ of $90\%$ pure limestone $(CaCO_3)$?

Similar Questions

The decomposition of $H_2O_2$ produces $O_2$. If at a certain time,$48 \ g$ of $O_2$ is produced in $1 \ minute$,then the rate of formation of water at that time will be .......... $mol \ min^{-1}$.

How many liters of $CO_2$ gas are released at $STP$ upon the complete decomposition of $9.85 \ g$ of $BaCO_3$ (Atomic mass of $Ba = 137$)?

The mass of $BaCO_3$ produced when excess $CO_2$ is bubbled through a solution of $0.205 \ mol$ $Ba(OH)_2$ is ............ $g$.

$120 \ g$ of urea are present in $5 \ L$ of solution. The active mass of urea is:

The weight of $H_2C_2O_4 \cdot 2H_2O$ required to prepare $500 \ mL$ of $0.2 \ N$ solution is .......... $g$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module