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How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na_2CO_3and NaHCO_3containing equimolar amounts of both ? |
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Answer» Solution :STEP 1. To calculate the number of moles of the components in the mixture. Suppose mass of `Na_2CO_3` PRESENT in the mixture = x g Mass of `NaHCO_3`present in the mixture = (1 – x) g Molar mass of `Na_2CO_3 = 2 xx 23 + 12 + 3 xx 16 = 106 g "mol"^(-1)` Molar mass of `NaHCO_3 = 23 + 1 + 12 + 3 xx 16 = 84 g "mol"^(-1)` Number of moles of `Na_2CO_3` in x g ` = (x)/(106)` Number of moles of `NaHCO_3` in (1 – x) g= `(1 - x)/(84)` As mixture CONTAINS EQUIMOLAR amounts of the two, `(x)/(106) = (1-x)/(84) " or " 106 - 106x = 84x " or " x= (106)/(106+ 84) = 106/190 g = 0.558 g ` Number of moles of `Na_2CO_3 = (0.558)/(106) = 0.00526` Number of moles of `NaHCO_3 = (1-0.558)/(84) = (0.442)/(84) = 0.00526` Step 2. To calculate the number of moles of HCl required. `Na_2CO_3 + 2HCl to 2NaCl + H_2O + CO_2` `NaHCO_3 + HCl to NaCl + H_2O + CO_2` `NaHCO_3 + HCl to NaCl + H_2O + CO_2` 1 mole of `Na_2CO_3` requires = 2 moles of HCI ` therefore ` 0.00526 mole of `Na_2CO_3`requires = 0.00526 x 2 moles = 0.01052 mole 1 mole of `NaHCO_3`requires = 1 mole of HCI ` therefore `0.0526 mole of `NaHCO_3` requires = 0.00526 mole ` therefore ` Total HCl required = 0.01052 + 0.00526 moles = 0.01578 moles Step 3. To calculate volume of 0.1 M HCl. 0.1 mole of 0.1 M HCl are present in 1000 mL of HCl. 0.01578 mole of 0.1 M HCl will be present in = `(1000)/(0.01) xx 0.01578 = 157.8 mL` of HCl |
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