Answer:
ΔH°rxn = -827.5 kJ
Explanation:
Let's consider the following balanced equation.
2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g)
We can calculate the standard enthalpy of reaction (ΔH°rxn) from the standard enthalpies of formation (ΔH°f) using the following expression.
ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g)
)] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g)
)]
ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g)
)] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g)
)]
ΔH°rxn = [2 mol × (-217.32 kJ/mol) + 2 mol × (-296.83)] - [2 mol × (-100.4) + 3 mol × 0 kJ/mol]
ΔH°rxn = -827.5 kJ
I think a controlled experiment
<span>c.
reactivity
this is a physical property depends on how reactive something is</span>
Answer:
Percent yield of PI3 = 95.4%
Explanation:
This is the reaction:
2P (s) + 3I2 (g) > 2PI3 (g)
Let's determine the moles of iodine that has reacted.
58.6 g / 253.8 g/mol = 0.231 mol
Ratio is 3:2. Let's make a rule of three to state the moles produced at 100 % yield reaction.
3 moles of I2 react to make 2 moles of PI3
0.231 moles of I2 would make (0.231 .2) / 3 = 0.154 moles of PI3
As we have produced 0.147 moles let's determine the percent yield.
(Yield produced / Theoretical yield) . 100 > (0.147 / 0.154) . 100 = 95.4%
Answer:
2,75 mol of O2 it's 88 g of O2.
Explanation:
The weight of the diatomic molecule O2 is 32 g/mol. So considering that, you should multiply 2,75 mol · 32 = 88g :)
