Answer:
31.58 L
Explanation:
From the question given above, the following data were obtained:
Initial mole (n₁) of gas = 1.90 moles
Initial volume (V₁) = 40 L
Final mole (n₂) = 1.90 – 0.40 = 1.5 moles
Final volume (V₂) =.?
The final volume of the gas can be obtained as follow;
V₁ / n₁ = V₂ / n₂
40 / 1.9 = V₂ / 1.5
Cross multiply
1.9 × V₂ = 40 × 1.5
1.9 × V₂ = 60
Divide both side by 1.9
V₂ = 60 / 1.9
V₂ = 31.58 L
Thus, the final volume of the gas is 31.58 L
1 Cal ---------- 4.184 J
? Cal ---------- 130.0 J
130.0 x 1 / 4.184 => 31.07 Cal
hope this helps!
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Answer:
= 97.44 Liters at S.T.P
Explanation:
The reaction between Iron (iii) oxide and Carbon monoxide is given by the equation;
Fe2O3(s)+ 3CO(g) → 3CO2(g) + 2Fe(s)
From the reaction when the reactants react, 2 moles of Fe and 3 moles of CO2 are produced.
Therefore; Mole ratio of Iron : Carbon dioxide is 2:3
Thus; Moles of Carbon dioxide = (2.9/2)×3
= 4.35 moles
But; 1 mole of CO2 at s.t.p occupies 22.4 liters
Therefore;
Mass of CO2 = 22.4 × 4.35 Moles
= 97.44 L
