Answer:
The grams that were removed are 0.787
Explanation:
We can raise two situations, with Ideal Gases Law.
First (S1): P. 1.15L =0.0353 . R .T
Second (S2): P. 0.790L = n . R .T
If we compare them, we must make a division, so we can know the moles of gas, that are still remained in the container.
S1/S2
P . 1.15L / P. 0.790L = 0.0353 . R.T / n . R. T
Pressure, temperature and R, are constant, so they can be cancelled.
So finally we have.
1.15L / 0.790L = 0.0353 moles / n
n = 0.0353 moles (0.790L /1.15L)
n = 0.0242 moles
Initially we have 0.0353 moles; afterwards we have 0.0242 moles. In the reaction we lost (0.0353 - 0.0242) = 0.0111 moles
The grams that were removed are:
Moles . molar mass = mass
0.0111 m . 70.9 g/m = 0.787 g
