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3 years ago

Calculate the volume occupiedat s.t.p by 6.89gas [H = 10, N = 14

Chemistry

1 answer:

Nadya [2.5K]3 years ago

8 0

Answer:

9.07 L

Explanation:

<em>Calculate the volume occupied at s.t.p by 6.89 g of NH₃ gas [H = 1.0, N = 14.0].</em>

Step 1: Given and required data

Mass of NH₃ (m): 6.89 g

Molar mass of NH₃ (M): 17.0 g/mol

Step 2: Calculate the moles (n) of NH₃

We will use the following epxression.

n = m / M

n = 6.89 g / (17.0 g/mol) = 0.405 mol

Step 3: Calculate the volume occupied by 0.405 moles of NH₃ at STP

At STP, 1 mole of NH₃ occupies 22.4 L (assuming ideal behavior).

0.405 mol × 22.4 L/1 mol = 9.07 L

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Partial pressure of the $PCl_3=p_1'=13.2 Torr$

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Partial pressure of the $PCl_5=p_3'=217.0 Torr$

Total pressure of the system = P = 263.0 Torr

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$263.0Torr=13.2 Torr+p_2'+217.0 Torr$

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(13.2) Torr (32.8) Torr (13.2) Torr

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(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr

$K_p=\frac{p_3'}{p_1'\times p_2'}$

$1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}$

Solving for x;

x = 6.402 Torr

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$p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr$

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Calculate the volume occupiedat s.t.p by 6.89gas [H = 10, N = 14​

Calculate the volume occupiedat s.t.p by 6.89gas [H = 10, N = 14