Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Saurabh on January 09, 2018, 08:01:22 AM
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150 mL of H2O2 sample was divided into two parts. First part was treated with KI and formed KOH required 200 mL of M/2 H2SO4 for neutralization. Other part was treated with KMnO4 yielding 6.74 litre of O2 at STP. Using % yield indicated find volume strength of H2O2 sample used.
H2O2 + 2KI ---------------> I2 + 2KOH
(40% yield)
H202 + 2KMnO4 + 3H2SO4 -----------> K2SO4 + 2MnSO4 + 3O2 + 4H20
(50% yield)
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Sarah,
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