AP EAMCET · Chemistry · Redox Reactions
On reduction with hydrogen, \(3.6 \mathrm{~g}\) of an oxide of metal \((M)\) left \(3.2 \mathrm{~g}\) of the metal. If the atomic weight of the metal is 64 . The formula of the oxide is
- A \(\mathrm{M}_2 \mathrm{O}_3\)
- B \(\mathrm{M}_2 \mathrm{O}\)
- C \(M O\)
- D \(\mathrm{MO}_2\)
Answer & Solution
Correct Answer
(B) \(\mathrm{M}_2 \mathrm{O}\)
Step-by-step Solution
Detailed explanation
Moles of \(M=\frac{3.2}{64}=\frac{1}{20} \mathrm{~mol}\) Weight of oxygen \(=3.6-3.2=0.4\) Moles of oxygen \(=\frac{0.4}{16}=\frac{1}{40}\)…
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