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2.0 g of mixure of Na2CO3, NaHCO3, and N...

2.0 g of mixure of `Na_2CO_3`, `NaHCO_3`, and NaCl on heating produced 56 " mL of " `CO_2` at STP. 1.6g of the same mixture required 25 " mL of " 0.5 M `H_2SO_4` for complete neutralisation. Calculate the percentage of each component present in the mixture.

Text Solution

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On heating only `NaHCO_3` decomposes to give `CO_2`
`2NaHCO_3toNa_2CO_3+H_2O+CO_2uarr`
" Eq of "`NaHCO_3=" Eq of "CO_2`
`(W_(NaHCO_3))/(Ew(NaHCO_3))=("Volume of "CO_2)/("Volume of "1 " Eq of "CO_2)`
`(W_(NaHCO_(3)))/((84)/(1))=(56)/((22400)/(2))`
`W_(NaHCO_3)` in `2.0g` of mixture `=(0.42)/(2)xx100`
`=21%` of mixture
Weight of `NaHCO_3` in 1.6 of mixture`=(1.6xx21)/(100)=0.336g`
Let the weight of `NaCl` be x g
Weight of `NaHCO_3=0.336g`
Weight of `Na_2CO_3=[1.6-(x+0.336)]g`
`=(1.264-x)g`
Since both `Na_2CO_3` and `NaHCO_3` react with `H_2SO_4`, `m" Eq of "Na_2CO_3+m" Eq of "NaHCO_3=m" Eq of "H_2SO_4`
`((1.264-x)/((126)/(2))+(0.336)/((84)/(1)))xx10^(3)=25xx0.5xx2` (n-factor)
`x=0.151 g`
Weight of `NaCl=0.151 g` and `%` of `NaCl=9.43%`
Weight of `NaHCO_3=0.336g` and `%` of `NaHCO_3=21.0%`
Weight of `Na_2CO_3=1.264-0.151=1.113`
and `%` of `Na_2CO_3=69.57%`
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