Posted by Fai on .
What weight of na2co3 of 95% purity would be required to neutralize 45.6ml of 0.235n acid?
I want you step by step clearly explain
mL acid x N acid = milliequivalents acid = 45.6 mL x 0.235 = ? m.e. acid
1 m.e. of anything = 1 m.e. of anything else; therefore, m.e. acid = m.e. Na2CO3.
?m.e. acid = ? m.e. Na2CO3.
m.e. Na2CO3 x milliequivalent weight Na2CO3 = grams Na2CO3
?m.e. Na2CO3 x 0.053 = grams Na2CO3.
Put all of this together to make
mL x N x mew = grams
45.6mL x 0.235N x 0.053g/m.e. =0.56795 g of 100% Na2CO3. Since it is only 95%, then
0.56795/0.95 = 0.5978g which rounds to 0.598g to three significant figures.