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2N2H4(l) + N2O4(l) → 3N2(g) + 4H2O(g) [balanced] How many moles of N2H4 is required to produce 28.3 g of N2? Assume that all rea

ctants react completely. Molar mass of N2H4 = 32.06 g/mol Molar mass of N2O4 = 92.02 g/mol Molar mass of N2 = 28.02 g/mol Molar mass of H2O = 18.02 g/mol Group of answer choices

1 answer:

JulijaS [17]1 year ago

6 0

Answer: 0.67 moles of $N_2H_4$

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles of nitrogen}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{28.3}{28.02}=1mole$

$2N_2H_4(l)+N_2O_4(l)\rightarrow 3N_2(g)+4H_2O(g)$

According to stoichiometry:

3 moles of $N_2$ is produced by 2 moles of $N_2H_4$

Thus 1 mole of $N_2$ is produced by= $\frac{2}{3}\times 1=0.67moles$ of $N_2H_4$

Thus 0.67 moles of $N_2H_4$ are required to produce 28.3 g of $N_2$

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Equilibrium: 0.1 - x x x

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$\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }$

$\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}$

$\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}$

<u>Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M</u>

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1 year ago