Correct option is (1) 1.76 g
Weight of impure limestone = 20 g
Weight of pure limestone (CaCO3) = 20% of 20 g
\(= \frac{20}{100} \times20\)
\(= 4g\)
\(n_{CaCO_3} = \frac 4{100} = 0.04\)
\(\underset{n = 0.04}{CaCO_3 }\to CaO +\underset{n = 0.04} {CO_2}\)
\(n_{CO_2} = 0.04\)
\(W_{CO_2} = 0.04 \times 44\)
\(= 1.76 g\)