Ap chem question?

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Ap chem question?

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Answer 1 · 2018-03-04T20:12:38

a.) $N H_{3}$ $NH_3$
b.) $P_{N H_{3}} = 1.80$ $P_(NH_3) = 1.80$ atm and $P_{H C l} = 0.838$ $P_(HCl) = 0.838$ atm
c.) $P = 0.960$ $P = 0.960$ atm
d.) It should be closer to $H C l$ $HCl$ 's chamber and the distance of $N H_{3}$ $NH_3$ should be about double that of $H C l$ $HCl$

Explanation:

a.) We want to know which one is limiting and which one is in excess for this. So we need to calculate moles of each reactant:

For $N H_{3}$ $NH_3$ :
$\frac{5}{17} = 0.294$ $5 / 17 = 0.294$ moles

For $H C l$ $HCl$ :
$\frac{5}{36.45} = 0.137$ $5/36.45 = 0.137$ moles

So there will be excess $N H_{3}$ $NH_3$ since $N H_{3}$ $NH_3$ and $H C l$ $HCl$ both have stoichiometric coefficients of 1.

b.) Let's use the ideal gas law for this, we want to find total pressure as the stopcock is opened, so:

$P = \frac{n R T}{V}$ $P = (nRT)/V$
$P = \frac{\frac{0.431}{1000} \cdot 82.05 \cdot (273 + 25)}{4}$ $P = (0.431/1000 * 82.05 * (273+25))/4$ The R constant is in Latm/kmolK so we have to adjust the moles to be in kmols (divide by 1000)

$P = 2.63$ $P = 2.63$ atm

Now partial pressures are dependent on mole fractions of each component, so the mole fraction of $H C l$ $HCl$ is $\frac{0.137}{0.431} = 0.318$ $0.137/0.431 = 0.318$ , multiply this by the total pressure to get $P_{H C l} = 0.838$ $P_(HCl) = 0.838$ atm. Subtract this from total pressure to get the partial pressure of $N H_{3}$ $NH_3$ , $P_{N H_{3}} = 1.80$ $P_(NH_3) = 1.80$ atm

c.) Again we can use ideal gas law, but some parameters have changed, as we lost moles of gas. Now it looks like this:

$P = \frac{\frac{0.157}{1000} \cdot 82.05 \cdot (273 + 25)}{4}$ $P = (0.157/1000 * 82.05 * (273+25))/4$
$P = 0.960$ $P = 0.960$ atm

The 0.157 came from the excess moles of $N H_{3}$ $NH_3$ , or those unreacted

d.) Ammonium will form where the gases meet, so we need to determine how much volume the gases will take up. We can use the pressures to estimate it, $N H_{3}$ $NH_3$ is pushing harder against $H C l$ $HCl$ so it should be closer to $H C l$ $HCl$ 's side. The pressure fraction can help approximate more accurately, so:

$\frac{P_{N H_{3}}}{P_{H C l}} = 2.148$ $P_(NH_3)/P_(HCl) = 2.148$

So in the line $N H_{3}$ $NH_3$ 's distance should be about twice $H C l$ $HCl$ 's distance, which is close to what you have marked up there.

Hope this helps!

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