Consider this reaction.2N2H4(l)+N2O4(l)⟶3N2(g)+4H2O(g) ΔH=−1078kJ How much energy is released by this reaction during the formation of 140 g of N2(g)?

2N2H4(g)⟶3N2(g)+4H2O Δ4=−1078kJ Mole N2=nN2=14028=5∴ΔH′ to 140 gm N210783×14028=1797kJ

Consider this reaction.2N2H4(l)+N2O4(l)⟶3N2(g)+4H2O(g) ΔH=−1078kJ How much energy is released by this reaction during the formation of 140 g of N2(g)?

2N2H4(g)⟶3N2(g)+4H2O Δ4=−1078kJ Mole N2=nN2=14028=5∴ΔH′ to 140 gm N210783×14028=1797kJ

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Consider this reaction.
$\begin{array}{r} 2 N_{2} H_{4} (l) + N_{2} O_{4} (l) ⟶ 3 N_{2} (g) + 4 H_{2} O (g) Δ H = - 1078 kJ \end{array}$
How much energy is released by this reaction during the formation of 140 g of $N_{2} (g)$ ?

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3234 kJ

5390 kJ

1078 kJ

1797 kJ

answer is B.

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Detailed Solution

$2 N_{2} H_{4} (g) ⟶ 3 N_{2} (g) + 4 H_{2} O Δ 4 = - 1078 kJ$

$Mole N_{2} = n_{N_{2}} = \frac{140}{28} = 5$

$∴ Δ H^{'} to 140 gm N_{2}$

$\frac{1078}{3} \times \frac{140}{28} = 1797 kJ$

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Consider this reaction.2N2H4(l)+N2O4(l)⟶3N2(g)+4H2O(g) ΔH=−1078kJ How much energy is released by this reaction during the formation of 140 g of N2(g)?

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Consider this reaction.
$\begin{array}{r} 2 N_{2} H_{4} (l) + N_{2} O_{4} (l) ⟶ 3 N_{2} (g) + 4 H_{2} O (g) Δ H = - 1078 kJ \end{array}$
How much energy is released by this reaction during the formation of 140 g of $N_{2} (g)$ ?