Question

what volume of O2 at S.T.P. will be produced on heating 65.6 g of Ca (NO3)2?​

Answer 1

4.48L

We have 65.6g, i.e 0.4 moles of Ca(NO3)2 with us. (65.6g/164g=0.4moles). Thus 0.4 moles of calcium nitrate gives 0.2 moles of dioxygen according to the balanced equation. Therefore, 0.2 moles of O2 corresponds to 4.48L at STP.