Answer: Since the reaction is 2Ca(NO3)2 = 2CaO + 4NO2 + O2
1) - given the stoichiometric coefficients, we know that 2 moles of Ca(NO3)2 will produce 4 moles of NO2, hence, 1 mole will produce 2 moles of NO2
2) - 328 g produces 22.4 L, since One mole of any gas at S.T.P. occupies the same volume which is 22.4 L.
Hence, 65.6 g produces (65.6*22.4)/328 = 4.48 L
3) - 328 g produces 112 g CaO. Therefore, 65.6 g produces = (65.6*112)/328 = 22.4 g CaO.
4) - Given the stoichiometric coefficients, we know 5 moles of gaseous products are already being produces ( 4+1) by 2 moles of reactant.
5) - 44.8 L at STP = 2 moles of NO2, since One mole of any gas at S.T.P. occupies the same volume which is 22.4 L.
Hence, to produce 2 moles of NO2, we need 1 mole of reactant = 164 g
Hope this helps :)
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