The number of molecules of Carbon(iv)Oxide produced when 10.0g of CaCO\(_3\) is treated with 0.2dm\(^3\)...

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 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\(_2\)

There is an explanation video available below.

{{ settings.no_comment_msg ? settings.no_comment_msg : 'There are no comments' }}