What is the total pressure in atmosphere (atm) of a gas mixture containing argon gas at 0.28 atm, helium gas at 760 mm Hg, and n
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Answer:- C. 16.4 L
Solution:- The given balanced equation is:
From this equation, there is 2:1 mol ratio between HCl and hydrogen gas. First of all we calculate the moles of hydrogen gas from given grams of HCl using stoichiometry and then the volume of hydrogen gas could be calculated using ideal gas law equation, PV = nRT.
Molar mass of HCl = 1.008 + 35.45 = 36.458 gram per mol
The calculations are shown below:
=
Now we will use ideal gas equation to calculate the volume.
n = 0.672 mol
T = 25 + 273 = 298 K
P = 101.3 kPa = 1 atm
R =
PV = nRT
1(V) = (0.672)(0.0821)(298)
V = 16.4 L
From calculations, 16.4 L of hydrogen gas are formed and so the correct choice is C.
<u>Answer:</u> The value of for given reaction is 0.465
<u>Explanation:</u>
We are given:
Initial concentration of A = 0.650 M
Initial concentration of B = 1.35 M
Initial concentration of C = 0.300 M
Equilibrium concentration of A = 0.550 M
Equilibrium concentration of B = 0.400 M
For the given chemical equation:
<u>Initial:</u> 0.65 1.35 0.30
<u>At eqllm:</u> 0.65-x 1.35-2x 0.30+x
Evaluating the value of 'x'
So, equilibrium concentration of B = 1.35 - 2x = [1.35 - 2(0.100)] = 1.15 M
Equilibrium concentration of C = (0.30 + x) = (0.300 + 0.100) = 0.400 M
The expression of for above equation follows:
Putting values in above equation, we get:
Hence, the value of for given reaction is 0.465