Answer:
n₂ =1.4 mol
Explanation:
Given data:
Mass of nitrogen = 2 g
Initial Volume occupy by nitrogen = 1.25 L
Final volume occupy by nitrogen = 25.0 L
Final number of moles = ?
Solution;
Formula:
V₁ / n₁ = V₂ / n₂
Number of moles of nitrogen:
Number of moles = mass/ molar mass
Number of moles = 2 g/ 28 g/mol
Number of moles = 0.07 mol
Now we will put the values in formula:
V₁ / n₁ = V₂ / n₂
n₂ = V₂× n₁ /V₁
n₂ = 25 L × 0.07 mol / 1.25 L
n₂ = 1.75 L. mol / 1.25 L
n₂ =1.4 mol
Answer:
ΔHorxn = - 11.79 KJ
Explanation:
2 SO 2 ( g ) + O 2 ( g ) ⟶ 2 SO 3 ( g )
The standard enthalpies of formation for SO 2 ( g ) and SO 3 ( g ) are Δ H ∘ f [ SO 2 ( g ) ] = − 296.8 kJ / mol Δ H ∘ f [ SO 3 ( g ) ] = − 395.7 kJ / mol
From the reaction above, 2 mol of SO2 reacts to produce 2 mol of SO3. Assuming ideal gas behaviour,
1 mol = 22.4l
x mol = 2.67l
Upon cross multiplication and solving for x;
x = 2.67 / 22.4 = 0.1192 mol
0.1192 mol of SO2 would react to produce 0.1192 mol of SO3.
Amount of heat is given as;
ΔHorxn = ∑mΔHof(products) − ∑nΔHof(reactants)
Because O2(g) is a pure element in its standard state, ΔHοf [O2(g)] = 0 kJ/mol.
ΔHorxn = 0.1192 mol * (− 395.7 kJ / mol) - 0.1192 mol * ( − 296.8 kJ / mol)
ΔHorxn = - 47.17kj + 35.38kj
ΔHorxn = - 11.79 KJ
