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How many mL of a 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 the two? |
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Answer» Solution :Step 1. To calculate the number of moles of the components in the mixture. Suppose `Na_(2)CO_(3)` present in the mixture = xg`""therefore""NaHCO_(3)" present in the mixture "=(1-x)g` `"Molar mass of "Na_(2)CO_(3)=2xx23+12+3xx16="106 g mol"^(-1)` `"Molar mass of "NaHCO_(3)=23+1+12+3xx16="84 g mol"^(-1)` `therefore"Moles of "Na_(2)CO_(3)" in x g "=(x)/(106),"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)/(190)g=0.558g` Thus,`"moles of "Na_(2)CO_(3)=(0.558)/(106)=0.00526,"Moles of "NaHCO_(3)=(1-0.558)/(84)=(0.442)/(84)=0.00526` Step 2. To calculate the moles of HCl required. `Na_(2)CO_(3)+2HCl rarr 2NaCl+H_(2)O+CO_(2),NaHCO_(3)+HCl rarr NaCl +H_(2)O+CO_(2)` `"1 mole of "Na_(2)CO_(3)" required HCl =2 moles"` `therefore" 0.00526 mole of "Na_(2)CO_(3)"REQUIRES HCl "=0.00526 xx"2 moles = 0.01052 mole"` `"1 mole of "NaHCO_(3)" requires HCl =1 mole"` `therefore"0.00526 mole of "NaHCO_(3)" required HCl"="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 HCl "=(1000)/(0.1)xx0.01578="157.8 mL"` |
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