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Question

Chemistry Question on Some basic concepts of chemistry

12 g of Mg with excess of HCI at NTP gives

A

\ce11.2dm3ofH2\ce{11.2 \,dm^3\, of\, H_2}

B

\ce22.4dm3ofH2\ce{22.4 \, dm^3\, of\, H_2 }

C

\ce5.6dm3ofH2\ce{5.6\, dm^3 \, of\, H_2 }

D

\ce15.6dm3ofH2\ce{ 15.6 \, dm^3 \, of\, H_2}

Answer

\ce11.2dm3ofH2\ce{11.2 \,dm^3\, of\, H_2}

Explanation

Solution

Mg+2HCl>MgCl2+H2{Mg+ 2HCl -> MgCl_2 + H_2}
(24g)(24\, g) (2gor22.4dm3) (2 \, g \,or\, 22.4\, dm^3)
Thus, 12 g of Mg produces= 1 g
or 11.2 dm3H2dm^3 \,H_2