100 g impure CaCO 3 on heating gives 5.6 L . CO 2 gas at NTP. Find the percentage of calcium in the limestone sample. [At. wt.: Ca =40, C =12, O =16 ]

Question

100 g impure CaCO 3 on heating gives 5.6 L . CO 2 gas at NTP. Find the percentage of calcium in the limestone sample. [At. wt.: Ca =40, C =12, O =16 ] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

CaCO 3 longrightarrow CaO + CO 2 (5.6/22.4)=(1/4) mol Mol of CaO = mol of Ca =(1/4) Mass of Ca =(1/4), × 40=10 % of Ca in sample =(10/100) × 100=10 %

Q. $100 g$ impure $C a C O_{3}$ on heating gives $5.6 L . C O_{2}$ gas at NTP. Find the percentage of calcium in the limestone sample. [At. wt.: $C a = 40, C = 12, O = 16$ ]

$C a C O_{3} ⟶ C a O + C O_{2}$
$\frac{5.6}{22.4} = \frac{1}{4} m o l$
Mol of $C a O = m o l$ of $C a = \frac{1}{4}$
Mass of $C a = \frac{1}{4}, \times 40 = 10$
$%$ of $C a$ in sample $= \frac{10}{100} \times 100 = 10%$

Q. 100g impure CaCO3​ on heating gives 5.6L.CO2​ gas at NTP. Find the percentage of calcium in the limestone sample. [At. wt.: Ca=40,C=12,O=16 ]

CaCO3​⟶CaO+CO2​ 22.45.6​=41​mol Mol of CaO=mol of Ca=41​ Mass of Ca=41​,×40=10 % of Ca in sample =10010​×100=10%

Q. $100 g$ impure $C a C O_{3}$ on heating gives $5.6 L . C O_{2}$ gas at NTP. Find the percentage of calcium in the limestone sample. [At. wt.: $C a = 40, C = 12, O = 16$ ]

$C a C O_{3} ⟶ C a O + C O_{2}$
$\frac{5.6}{22.4} = \frac{1}{4} m o l$
Mol of $C a O = m o l$ of $C a = \frac{1}{4}$
Mass of $C a = \frac{1}{4}, \times 40 = 10$
$%$ of $C a$ in sample $= \frac{10}{100} \times 100 = 10%$