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what is the normality and nature of the mixture obtained by mixing 0.62 gram of na2co3.h2o to 100 ml of 0.1N h2so4Asked by jaimi
What is the normality and nature of a mixture obtained by mixing 0.62 g of Na2CO3 •H2O to 100mL of 0.1N H2SO4?
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Answered by
DrBob222
I don't understand the problem.
Na2CO3 + H2SO4 ==> Na2SO4 + H2O + CO2
There is the equation. I have no idea what "nature of the mixture" means but let's see what happens.
How many grams Na2CO3 will the H2SO4 use? That's ml x N x milliequivalent weight Na2CO3 = grams Na2CO3.
100 mL x 0.1N x 0.053 = approx 0.52 but you need a more accurate answer than that.
You had 0.62 g Na2CO3 initially, you have used 0.52 which leaves about 0.1 g Na2CO3 unused. (Remember to recalculate all of these numbers.) The normality of that solution will be N = eq/L
equivalents = grams/eq weight = 0.1/53 = ?
N = eq/L = ?eq/0.1L = ?
So the nature of the solution may be that the H2SO4 is completely used and you have a solution of Na2CO3 in the liquid and the N of the Na2CO3 solution is ?? N.
Na2CO3 + H2SO4 ==> Na2SO4 + H2O + CO2
There is the equation. I have no idea what "nature of the mixture" means but let's see what happens.
How many grams Na2CO3 will the H2SO4 use? That's ml x N x milliequivalent weight Na2CO3 = grams Na2CO3.
100 mL x 0.1N x 0.053 = approx 0.52 but you need a more accurate answer than that.
You had 0.62 g Na2CO3 initially, you have used 0.52 which leaves about 0.1 g Na2CO3 unused. (Remember to recalculate all of these numbers.) The normality of that solution will be N = eq/L
equivalents = grams/eq weight = 0.1/53 = ?
N = eq/L = ?eq/0.1L = ?
So the nature of the solution may be that the H2SO4 is completely used and you have a solution of Na2CO3 in the liquid and the N of the Na2CO3 solution is ?? N.
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