Given 7 equiv calcium atoms; 10 equiv of carbon atoms; and 9 equiv of nitrogen atoms, how many equiv of #"calcium cyanide"# could we make?

1 Answer
anor277
Apr 26, 2017

The formula for #"calcium cyanide"# is #Ca(C-=N)_2#; we could make four such formula units with the given quantities.

Explanation:

#"Calcium cyanide"# is formulated by taking the calcium ion, #Ca^(2+)#, and binding it to 2 equiv of the cyanide ion, #""^(-):C-=N#.

So for #7xxCa#, #10xxC#, and #9xxN#, at most we could make #9xx^(-):C-=N# anions.

However from these anions, we could make AT MOST only #4xxCa(C-=N)_2# units. Are you happy with this?

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