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Nezavi [6.7K]
3 years ago
9

You have an unknown quantity of oxygen at a pressure of 2.2 atam, a volume of 21 liters and a temperature of 87 Celsius. How man

y miles of gas are left for you to breath
Chemistry
1 answer:
laila [671]3 years ago
6 0

<span>Let's assume that the oxygen gas has ideal gas behavior. 
Then we can use ideal gas formula,
      PV = nRT</span>


Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin.

<span>
P = 2.2 atm = 222915 Pa
V = 21 L = 21 x 10</span>⁻³ m³

n = ?

R = 8.314 J mol⁻¹ K⁻¹

<span> T = 87 °C = 360 K

By substitution,
</span>222915 Pa x 21 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻<span>¹ x 360 K
                                       n = 1.56</span><span> mol</span>

<span>
Hence, 1.56 moles of the oxygen gas are </span><span>left for you to breath.</span><span>
</span>
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Answer:

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In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

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Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\  n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%

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