A→2C+D3A→5E+2F
In reaction (i),
10 moles of C was formed.
As per the reaction,
2 moles of C is produced by 1 mole of A.
10 moles of C will be produced by 5 moles of A.
Amount of A left for 2nd reaction,
=15−5=10 mol
In reaction (ii),
3 moles of A produce 5 moles of E.
10 moles of A will produce,
= 53×10=16.67 mol