The number of molecules of Carbon(iv)Oxide produced when ...
The number of molecules of Carbon(iv)Oxide produced when 10.0g of CaCO3 is treated with 0.2dm3 of 1 Mole of HCL in the equation
CaCO3 + 2HCL ⇒ CaCl2 + H2 O + CO2 , is?
1.00 X 1023
6.02 X 1023
6.02 X 1022
6.02 X 1024
Correct answer is C
1 mole of CaCO3 = 100g or 6.02 X 1023
liberated 1 mole of CO2 or 44g or 6.02 X 1023
:100g = 6.02 X 1023
10g of CaCO3 = x
cross multiply
[10g X 6.02 X 1023] / 100g = x
⇒ x = 6.02 X 1022
Since
6.02 X 1023 of CaCO3 liberated 6.02 X 1023 of CO2
⇒ 6.02 X 1022 of CaCO3 will liberate 6.02 X 1022 of CO2
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