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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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2 years ago
A sealed 1.0 L flask is charged with 0.500 mol of I2 and 0.500 mol of Br2. An equilibrium reaction ensues: I2 (g) + Br2 (g) ↔ 2I
Daniel [21]

Answer:  Thus the value of K_{eq} is 110.25

Explanation:

Initial moles of  I_2 = 0.500 mole

Initial moles of  Br_2 = 0.500 mole

Volume of container = 1 L

Initial concentration of I_2=\frac{moles}{volume}=\frac{0.500moles}{1L}=0.500M

Initial concentration of Br_2=\frac{moles}{volume}=\frac{0.500moles}{1L}=0.500M

equilibrium concentration of IBr=\frac{moles}{volume}=\frac{0.84mole}{1L}=0.84M [/tex]

The given balanced equilibrium reaction is,

                            I_2(g)+Br_2(g)\rightleftharpoons 2IBr(g)

Initial conc.              0.500 M     0.500 M             0  M

At eqm. conc.    (0.500-x) M      (0.500-x) M     (2x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[IBr]^2}{[Br_2]\times [I_2]}

K_c=\frac{(2x)^2}{(0.500-x)\times (0.500-x)}

we are given : 2x = 0.84 M

x= 0.42

Now put all the given values in this expression, we get :

K_c=\frac{(0.84)^2}{(0.500-0.42)\times (0.500-0.42)}

K_c=110.25

Thus the value of the equilibrium constant is 110.25

8 0
3 years ago
What is a change that will not affect the pressure in a container?​
allochka39001 [22]

Answer:

Density of the contents

Explanation:

7 0
3 years ago
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