Question

If 20.0 g of CaCO3 is treated with 20.0g of HCl, how many gram of CO2 will be produced ?

Answer 1

Solution :

CaCO31 mol100g+2HCl(aq)2 mol73g→CaCl2(aq.)+H2O(l)+CO2g1 mol44g

Let CaCO3(s) be completely consumed in the reaction.

∴100gCaCO3give44gCO2

∴20gCaCO3will give44100×20gCO2=8.8CO2

Let HCl be completely consumed.

∴73gHCl give44gCO2

∴20g of HCl will give4473×20gCO2=12.054gCO2

Since, CaCO3 gives least amount of product CO2, hence, CaCO3 is limiting reactant. Amount of CO2 formed will be 8.8g

Answer 2

Answer:

So the limiting reagent is CaCO3. therefore 20 g of CaCO3 will give = 44/100 *20 = 8.8 g CO2.

Explanation: