1.

CaCO_3 + 2HCl to CaCl_2 + H_2O + CO_2 The volume of CO_2 gas formed when 2.5 g calcium carbonate are dissolved in excess hydrochloric acid at 0^@C and 1 atm pressure is [1 mole of any gas at 0^@Cand 1 atm pressure occupies 22.4141 volume]

Answer»

1.12L
56.0 L
0.28L
0.56L

Solution :`100 g CaCO_3 to 44g CO_2`
`2.5g CaCO_3` will liberate `44/100 xx 2.5g ` of `CO_2`
1 mole of `CO_2 = 44g`
no. of moles of `CO_2` = mass of `CO_2`/molar mass of `CO_2`` = ((44 xx 2.5)/(100) )/(44)`
` = (2.5)/(100) = 0.025` moles
1 mole of `CO_2`occupies 22.4L
` THEREFORE ` 0.025 moles of `CO_2` will OCCUPY `22.4 xx 0.025 = 0.56 L`


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