For the following reaction, if i start this reaction with 153.0 grams of hydrobromic acid and an excess of potassium bi sulfate, a) how many grams of sulfur dioxide gas will I produce? b) if 98.75 g are actually achieved, what is the % yield of the reaction? HBr(aq) + KHSO3(aq) --> KBr(aq)+SO2(g)+H2O(I)
a)
Balanced reaction
HBr (aq) + KHSO3 (aq) KBr (aq) + SO2 (g) + H2O (g)
mole ratio of HBr and SO2 = 1:1
now moles of HBr used
= mass of HBr / molar mass of HBr
= 153.0 (g)
= 1.89
therefore moles of SO2 can be formed = 1.89
mass of SO2 = moles* molar mass
= 1.89 (mol) 64 ( )
= 121.02 g
b)
Percent yield
=
=
= 81.59 %
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