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How many ml of 0.1 HCl is required to react completely with 1.0g mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` containing equi-molar amounts of both ? |
Answer» Given 1gm mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` Let the mass of `Na_(2)CO_(3)=" a gms"` Mass of `"NaHCO"_(3)=(1-a)" gms"` Number of moles of `"Na"_(2)"CO"_(3)=("wt")/("GMW")=(a)/(106)` Number of moles of `"NaHCO"_(3)=("wt")/("GMW")=(1-a)/(84)` Given that the mixture contains Equi molar amounts of `"Na"_(2)"CO"_(3)` and `"NaHCO"_(3)` `:. (a)/(106)=(1-a)/(84)` `84" a"=106-106" a"` `190"a"=106` a = 0.558 gms `:." Weight of "Na_(2)CO_(3)=0.558" gms"` Weight of `"NaHCO"_(3)=1-0.558=0.442" gms"` `"Na"_(2)"CO"_(3)+2" HCl"to2" NaCl"+"H"_(2)"O"+"CO"_(2)` 106 gms - 73 gms `0.558"g"-?=(73xx0.558)/(106)=0.384" gms"` `"NaHCO"_(3)+"HCl" to" NaCl"+"H"_(2)"O"+"CO"_(2)` 84 gms - 36.5 gms 0.442gms - ? `=(36.5xx0.442)/(84)=0.1928` `:." The weight of HCl required "=0.384+0.192=0.576" gms"` Molarity (M) `=("Wt")/("GMWt")xx(1000)/("V"(ml))` `0.1=(0.576)/(36.5)xx(1000)/("V")` `V=(0.576xx1000)/(36.5xx0.1)=(576)/(3.65)=157.80" ml"` |
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