When 10 g of 90% pure limestone is heated, the approximate volume (in L) of CO_2 liberated at STP is

Question

When 10 g of 90% pure limestone is heated, the approximate volume (in L) of CO_2 liberated at STP is:

MARKS Content Team · Accepted Answer

Correct answer: (B) 2.0. The decomposition reaction of CaCO_3 is, CaCO3_( (s))⟶∆CaO_((s))+CO2_( (g)) At STP, heating 1 mole of CaCO_3 (i.e., 100 g) liberates 1 mole or 22.4 L of CO_2. 10 g of 90% pure limestone=(10×90)/100 =9 g of pure CaCO_3. Since, 100 g→22.4 L Therefore, 9 g of CaCO_3 will liberate =(9×22.4)/100L =2.0 L CO_2

When $10 g$ of $90 %$ pure limestone is heated, the approximate volume (in $L$ ) of ${CO}_{2}$ liberated at $STP$ is:

Solution

Relevant Questions

When 10 g of 90% pure limestone is heated, the approximate volume (in L) of CO2 liberated at STP is:

Solution

Relevant Questions

When $10 g$ of $90 %$ pure limestone is heated, the approximate volume (in $L$ ) of ${CO}_{2}$ liberated at $STP$ is: