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How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na_(2)CO_(3) and NaHCO_(3) containing equimolar amounts of both ? |
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Answer» Solution :Let the amount of `Na_(2)CO_(3)` in the MIXTURE be `x_(g)`, Then, the amount of `NaHCO_(3)` in the mixture is `(1-x)g`. MOLAR mass of `Na_(2)CO_(3)=2xx23+1xx12+3xx16` `= 106 g mol^(-1)` `therefore` Number of moles `Na_(2)CO_(3)=(x)/(106)mol` Molar mass of `NaHCO_(3)=1xx23+1xx12+3xx16` `= 84 g mol^(-1)` Number of moles `NaHCO_(3)=(1-x)/(84)` mol According to the question, `(x)/(106)=(1-x)/(84)` `therefore 84x = 106-106 x` `therefore 190x=106` `therefore x=0.5579` Therefore, number of moles of `Na_(2)CO_(3)` `=(0.5579)/(106)mol` = 0.053 mol And, number of moles of `NaHCO_(3)` `= (1-0.5579)/(84)` = 0.0053 mol HCl reacts with `Na_(2)CO_(3)` and `NaHCO_(3)` according to the following equation. `2HCl+Na_(2)CO_(3)to 2NaCl+H_(2)O+CO_(2)` `HCl+NaHCO_(3)to NaCl+H_(2)O+CO_(2)` 1 mol of `Na_(2)CO_(3)` reacts with 2 mol of HCl. Therefore, 0.0053 mol of `Na_(2)CO_(3)` reacts with `2xx0.0053` mol = 0.0106 mol of HCl. Similarly, 1 mol of `NaHCO_(3)` reacts with 1 mol of HCl.Therefore, 0.0053 mol of `NaHCO_(3)` reacts with 0.0053 mol of HCl. Total moles of HCl required `= (0.0106+0.0053)` mol = 0.0159 mol In 0.1 M of HCl 0.1 mol of HCl is preset in 1000 mL of the solution Therefore, 0.0159 mol of HCl is present in `= (1000xx0.0159)/(0.1)` mol = 159 mL of the solution Hence, 159 mL of 0.1 M of HCl is required to react completely with 1 g mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` containing equimolar amounts of both. |
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