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?
A
1.00 X 10\(^{23}\)
B
6.02 X 10\(^{23}\)
C
6.02 X 10\(^{22}\)
D
6.02 X 10\(^{24}\)
Verified Correct Answer: C.
Chemical Principles & Mechanism
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\(_2\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #56769)