If 41.38n mL of a sulfuric acid solution reacts with 0.2888g of sodium carbonate dissolved in sufficient water the molarity of the sulfuric acid solution is?
Answer:
The equation is :
H2SO4 + Na2CO3 >> H2CO3 + Na2SO4
molecular weight Na2CO3 = 106 g/mol
0.2888 g / 106 = 0.002724 moles Na2CO3
The ratio between H2SO4 and Na2CO3 is 1 : 1.
Moles H2SO4 = 0.002724
M = 0.002724 mol / 0.04138 L = 0.06583 M
0.2888g Na2CO3 (MW=105.98874g) is 0.002725 mol
The reaction stoichiometry is 1:1.
H2SO4 + Na2CO3 ~> Na2SO4 + 2H2O
0.04138 L * (acid molarity) = 0.002725 mol
acid molarity = 0.06585 M
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