From 160g of so2 1.2×10^24 so2 are removed calculate the volume in liter at stp of remaining so2 gases
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No. of moles present in 160g SO2 =160/64= 2.5 moles. So after the removal of 1.204 x 10^ 24 molecules (2 moles) the remaining moles = (2.5–2.0)= 0.5. The volume of 0.5 mole SO2 at STP =22.4 x 0.5 = 11.2 Litres.
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