Question

100ml of solution containing 18.25g of HCL was heated 10 230 molecules were lost. The number of HCL molecules left ..​

Answer 1

Answer:

8.8 gram

Explanation:

100g

1mol

CaCO  3 (s) +  73g 2mol 2HCl(aq.)  →CaCl  2  (aq.)+H  2 ​  O(l)+  44g 1mol CO  2 (g)

Let CaCO  3  (s) be completely consumed in the reaction.  ∵100gCaCO  3

​   give 44gCO  2

​   ∴20gCaCO  3  will give   100 44  ×20gCO  2

​  =8.8gCO  2

​ Let HCl be completely consumed.

∵ 73 g HCl give 44 g CO2

​ ∴ 20 gHCl  will give   73( 44)  ×20gCO  2

​ =12.054gCO  2

​ Since, CaCO  3

​  gives least amount of product CO  2

​  , hence CaCO  3  is limiting reactant. Amount of CO  2

​ formed will 8.8 g.