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\(^{24}\)
jamb 2020chemistry

The correct answer is C. 6.02 X 10\(^{22}\)

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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