Asked by Fai
0.5g of fuming h2s04 oleum is diluted with water. This solution is completely neutralized by 26.7ml of 0.4N naoh. Find the percentage of free so3 in the sample
My calculation
h2s04 = 98g/mol
s03 = 80g/mol
98/2 = 49
80/2 = 40
26.7ml x 0.4n = 10.68ml - 0.01067L
x/49 + x/40 (0.5-x) = 0.01068L Please help me The answer 20.6%
My calculation
h2s04 = 98g/mol
s03 = 80g/mol
98/2 = 49
80/2 = 40
26.7ml x 0.4n = 10.68ml - 0.01067L
x/49 + x/40 (0.5-x) = 0.01068L Please help me The answer 20.6%
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