Posted by Fai on Sunday, January 27, 2013 at 2:39pm.
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.
Related Questions
chemistry - what weight of na2c03 of 95% purity would be required to neutralize ...
Gen Chem 1 - A tanker truck carrying 2.01×10^3kg of concentrated sulfuric ...
chemistry - 6.0g of impure and hydrous Na2SO4 were made up to 1dm^3 of solution ...
chemistry - 6.0g of impure and hydrous Na2SO4 were made up to 1dm^3 of solution ...
Chemistry - How do you find the concentration of an acid if 35.0 mL of 0.20 M ...
Chemistry - A 8.71 g sample of an aqueous solution of hydroiodic acid contains ...
Chemistry - A sample of oxalic acid, H2C2O4, is titrated with standard sodium ...
Chemistry - Acid spills are often neutralized with sodium carbonate. For example...
chemistry - 1. if 15.0mL of 4.5 M NaOH are diluted with water to a volume of ...
chemistry - How many grams of Na2CO3 are required to completely react with 35....
For Further Reading
