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