answer please
Q.(i) A gas of mass 32 gms has a volume of 20 litres at S.T.P. Calculate the gram molecular weight of the gas.
(ii) How much Calcium oxide is formed when 82 g of calcium nitrate is heated? Also find the volume of nitrogendioxide evolved:
(Ca = 40, N = 14, O = 16)
Dear Student,
i)
ii)
The reaction is 2Ca(NO3)2 = 2CaO + 4NO2 + O2
Here 2 moles of Ca(NO3)2 will produce 4 moles of NO2, hence, 1 mole will produce 2 moles of NO2
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, 82 g produces (82*22.4)/328 = 5.6 L
328 g produces 112 g CaO. Therefore, 82 g produces = (82*112)/328 = 28 g CaO.
We know 5 moles of gaseous products are already being produces ( 4+1) by 2 moles of reactant.
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.
So 82 gram reactant form 1 mole NO2
​Regards
i)
ii)
The reaction is 2Ca(NO3)2 = 2CaO + 4NO2 + O2
Here 2 moles of Ca(NO3)2 will produce 4 moles of NO2, hence, 1 mole will produce 2 moles of NO2
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, 82 g produces (82*22.4)/328 = 5.6 L
328 g produces 112 g CaO. Therefore, 82 g produces = (82*112)/328 = 28 g CaO.
We know 5 moles of gaseous products are already being produces ( 4+1) by 2 moles of reactant.
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.
So 82 gram reactant form 1 mole NO2
​Regards