part B
mass of NaCl produced = 7.976 - 7.625 = 0.351 gram
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we know the reaction of preparation of NaCl from Na2CO3 is Na2CO3 + 2 HCl -----> H2CO3 + 2NaCl
from this reaction we get to know that
105.986 gm of Na2CO3 produces = 2x 58.442 gram of NaCl hence 1 gm NaCl needs = 109.986 / 116.882 = 0.907 gm
hence 0.351 gm NaCl produced by = 0.907 x 0.351 = 0.318 gm
now we know 1 mole of Na2CO3 = 105.986 gm
or we can say 105.986 gm of Na2CO3 = 1 mole
this says 1 gm = 1/105.986 gm
hence 0.318 gm = [1 / 105.986] x 0.318 = 0.003 moles of Na2CO3
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molarity of Na2CO3 = moles of Na2CO3 / volume in liter = 0.003 moles / 0.005 liter = 0.6 M is molarity
--------part C -----
mass of unknown solid = mass of beaker with unknown - mass of beaker only
= 8.387 - 8.125 = 0.262 gram
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mass of NaCl produced = ,mass of beaker with NaCl - mass of beaker only
= 8.405 - 8.125 = 0.280 grams
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theoretical yield of NaCl will be calculated by eqation
Na2CO3 + 2 HCl -----> H2CO3 + 2 NaCl given that unknown solid is Na2CO3 [ 0.262 gram]
105.986 gram Na2CO3 produces = 2x58.441 = 116.882 gram NaCl
hence 0.280 gram of Na2CO3 will produce = [116.882/105.986]x0.262 = 0.289 grams of NaCl theoretically
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if unknown solid is NaHCO3
the eqations belongs to it
NaHCO3 + HCl -----> NaCl + H2CO3 this eqation says 84.005 gram of NaHCO3 produces = 58.441 grams of NaCl
hence 0.262 gram of NaHCO3 will produce = [58.441/84.005]x0.262 = 0.182 gram of NaCl produced
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