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3 years ago

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In the reaction :N2 + 3H2----2NH3 how many grams of H2 are needed to produce 1 mole on ammonia

1 answer:

svetlana [45]3 years ago

3 0

Start with moles of ammonia you want to produce, then convert to how many moles of H2 you need to get one mole of ammonia, then convert that to grams :)

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1.00 L of a gas at STP is compressed to 473mL. What is the new pressure of gas?

Ratling [72]

Hello!

1.00 L of a gas at STP is compressed to 473 mL. What is the new pressure of gas?

<u><em>We have the following data:</em></u>

Vo (initial volume) = 1.00 L

V (final volume) = 473 mL → 0.473 L

Po (initial pressure) = 1 atm (pressure exerted by the atmosphere - in STP)

P (final pressure) = ? (in atm)

<u><em>We have an isothermal transformation, that is, its temperature remains constant, if the volume of the gas in the container decreases, so its pressure increases. Applying the data to the equation Boyle-Mariotte, we have:</em></u>

$P_0*V_0 = P*V$

$1*1 = P*0.473$

$1 = 0.473\:P$

$0.473\:P = 1$

$P = \dfrac{1}{0.473}$

$\boxed{\boxed{P \approx 2.11\:atm}}\:\:\:\:\:\:\bf\green{\checkmark}$

<u><em>Answer: </em></u>

<u><em>The new pressure of the gas is 2.11 atm </em></u>

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$\bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}$

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A tank with volume of 2.4 cu ft is filled with Methane to a pressure of 1500 psia at 104 degrees F. Determine the molecular weig

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Explanation:

It is known that equation for ideal gas is as follows.

PV = nRT

The given data is as follows.

Pressure, P = 1500 psia, Temperature, T = $104^{o}F$ = 104 + 460 = 564 R

Volume, V = 2.4 cubic ft, R = 10.73 $psia ft^{3}/lb mol R$

Also, we know that number of moles is equal to mass divided by molar mass of the gas.

n = $\frac{mass}{\text{molar mass}}$

m = $n \times W$

= $0.594 \times 16.04$

= 9.54 lb

Hence, molecular weight of the gas is 9.54 lb.

We will calculate the density as follows.

d = $\frac{PM}{RT}$

= $\frac{1500 \times 16.04}{10.73 \times 564}$

= 3.975 $lb/ft^{3}$

Now, calculate the specific gravity of the gas as follows.

Specific gravity relative to air = $\frac{\text{density of methane}}{\text{density of air}}$

= $\frac{3.975 lb/ft^{3}}{0.0765 lb/ft^{3}}$

= 51.96

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