The number of molecules of Carbon(iv)Oxide produced when 10.0g of CaCO\(_3\) is treated with 0.2dm\(^3\) of 1 Mole of HCL in the equation
CaCO\(_3\) + 2HCL ⇒ CaCl\(_2\) + H\(_2\) O + CO\(_2\) , is?
- 1.00 X 10\(^{23}\)
- 6.02 X 10\(^{23}\)
-
6.02 X 10\(^{22}\)
- 6.02 X 10\(^{24}\)
The correct answer is: C
Explanation
1 mole of CaCO\(_3\) = 100g or 6.02 X 10\(^{23}\)
liberated 1 mole of CO\(_2\) or 44g or 6.02 X 10\(^{23}\)
:100g = 6.02 X 10\(^{23}\)
10g of CaCO\(_3\) = x
cross multiply
[10g X 6.02 X 10\(^{23}\)] / 100g = x
⇒ x = 6.02 X 10\(^{22}\)
Since
6.02 X 10\(^{23}\) of CaCO\(_3\) liberated 6.02 X 10\(^{23}\) of CO\(_2\)
⇒ 6.02 X 10\(^{22}\) of CaCO\(_3\) will liberate 6.02 X 10\(^{22}\) of CO\(_2\)
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