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A metal oxide has the formula `A_(2)O_(3)`. It can be reduced by hydrogen to give free metal and water. 0.1596 g of this metal oxide requires 6 mg of hydrogen for complete reduction. What is the atomic wight of metal?A. `52.3`B. `57.3`C. `55.8`D. `59.3` |
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Answer» Correct Answer - C `underset(0.1596 g)(A_(2) O_(3) )+ underset(0.006 g)(3 H_(2)) rarr 2 A + 3H_(2) O` `0.006 " g of" H_(2)` reduces ` 0. 1596 ` g of ` A_(2) O_(3)` 6 g og `H_(2)` will reduce `(0.1596xx6)/(0.006) = 159.6" g of " A_(2) O_(3)` Thus, molar mass of `A_(2) O_(3) = 159.6` g Let, atomic weight of A = x `therefore 2x + 3 xx 16 = 159.6 ` `2=159.6 - 48 = 111.6` `x = 55.8` |
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