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`        A 1g sample of KClO3 was heated under such conditions that a part of it decomposed according to the following equation:2KClO3 -> 2KCl+3O2The remaining part underwent change according to the following equation:4KClO3 -> 3KClO4+KClIf amount of O2 evolved was 146.8ml at NTP, calculate the percent by mass of KClO4 in the residue.`
5 months ago

```							2 KClO3 ? 2 KCl + 3 O2(0.1458 L O2) / (22.414 L/mol) x (2/3) x (122.5498 g KClO3/mol) = 0.5314 g KClO3 participated in the first reaction 1 g - 0.5314 g = 0.4686 g KClO3 decomposed according to the second reaction4 KClO3 ? 3 KClO4 + KCl(0.4686 g KClO3) / (122.5498 g KClO3/mol) x (3/4) x (138.5492 g KClO4/mol) = 0.3973 g KClO4 formed in the second reaction
```
5 months ago
```							Dear student The above ans is incorrectMoles of   KClO3 = 1/122.55 g/mol = 0.081 mol(1) 2KClO3 - 2KCl + 3O2 Moles of oxygen gas at 1 atm and 273 K can be calculated by using ideal gas equation :n = PV/RT n = 0.00499 mol = Moles of oxygen gas produced at given conditionsAccording to reaction(1), 2 moles of  produces 3 moles of oxygen.Then 0.00499 mol of oxygen will be produced from: of 2/3 * 0.0499 mol = 0.00332 mol of KClO3 Moles of  used in part(1) = 0.00332 mol(2) 4KClO3 - 3KclO4 +Kcl Moles of  used in part(2)== 0.0081 mol - 0.00332 mol = 0.00478 molAccording to reaction (2), 4 moles of KClO3  gives 3 moles of Kcl O4  .Then 0.00478 moles of KClO3  will give: 3/4 * 0.00478 mol = 0.003585 mol of KCl O3Mass of 0.003585 moles of:0.003585 mole × 138.55 g/mol =0.4967 gPercentage by weight of  in the residue:% = Mass of KClO4 / Mass of KClO3= 49% The % by weight of  in the residue 49.67 %
```
5 months ago
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