4. A student started with a 0.032 g sample of copper which he took through the series of reactions described in this experiment.
1 answer:
Answer:
Explanation:
Hello,
In this case, we can consider the following chemical reaction for the oxidation of copper which only occurs at high temperatures:
In such a way, for 0.032 grams of copper, the following grams of copper (II) oxide (black product) are yielded:
Therefore, the percent yield is:
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Answer:
Therefore, the amount of heat produced by the reaction of 42.8 g S = <u>(-5.2965 × 10²) kJ = (-5.2965 × 10⁵) J</u>
Explanation:
Given reaction: 2S + 3O₂ → 2 SO₃
Given: The enthalpy of reaction: ΔH = - 792 kJ
Given mass of S: w₂ = 42.8 g, Molar mass of S: m = 32 g/mol
In the given reaction, the number of moles of S reacting: n = 2
As, Number of moles:
∴ mass of S in 2 moles of S:
<em>Given reaction</em>: 2S + 3O₂ → 2 SO₃
<em>In this reaction, the limiting reagent is S</em>
⇒ 2 moles S produces (- 792 kJ) heat.
or, 64 g of S produces (- 792 kJ) heat.
∴ 42.8 g of S produces (x) amount of heat
⇒ <u><em>The amount of heat produced by 42.8 g S:</em></u>
<u>Therefore, the amount of heat produced by the reaction of 42.8 g S = (-5.2965 × 10²) kJ = (-5.2965 × 10⁵) J</u>