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Aleksandr-060686 [28]
3 years ago
9

New concept cars in the Detroit Auto Show feature engines that burn hydrogen gas in air to produce water vapor. Suppose that fue

l tank contains 150. L of H2 gas at 20.0 atm pressure at 25.0˚C. If all of this gas is burned, what mass (in grams) of water vapor is produced?
Chemistry
1 answer:
Maksim231197 [3]3 years ago
3 0

Answer:

247.2g

Explanation:

Using the general gas law equation as follows:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

According to the information in this question, a fuel tank contains 150. L (V) of H2 gas at 20.0 atm (P) at 25.0˚C (T)

Temperature = 25°C = 25 + 273 = 298K

Using PV = nRT

20 × 150 = n × 0.0821 × 298

3000 = 24.4658n

n = 3000/24.4658

n = 122.62

n = 122.6mol

Using the formula, mole = mass/molar mass, to find the mass of H2 gas.

Molar mass of H2 = 1.008(2)

= 2.016g/mol

122.6 = mass/2.016

mass = 122.6 × 2.016

mass = 247.2g.

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The answer for this issue is: 
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3 0
4 years ago
A 720. cm^3 vessel contains a mixture of Ar and Xe. If the mass of the gas mixture is 2.966 g at 25.0°C and the pressure is 760.
sleet_krkn [62]

Explanation:

The given data is as follows.

      Pressure (P) = 760 torr = 1 atm

      Volume (V) = 720 cm^{3} = 0.720 L

     Temperature (T) = 25^{o}C = (25 + 273) K = 298 K

Using ideal gas equation, we will calculate the number of moles as follows.

                                PV = nRT

   Total atoms present (n) = \frac{PV}{RT}

                                          = 1 \times \frac{0.720 L}{0.0821 \times 298}

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Let us assume that there are x mol of Ar and y mol of Xe.

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               x + y = 0.0294

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                = 0.0197 mol

Therefore, mole fraction will be calculated as follows.

Mol fraction of Xe = \frac{y}{(x+y)}

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Answer:

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