Answer:
20 molecules of oxygen gas remains after the reaction.
Explanation:

Molecules of ethyne = 52
Molecules of oxygen gas = 150
According to reaction, 2 molecules of ethyne reacts with 5 molecules of oxygen gas.
Then 52 molecules of ethyne will react with:
of oxygen gas.
As we can see that we have 150 molecules of oxygen gas, but 52 molecules of ethyne will react with 130 molecules of oxygen gas. So, this means that ethyne is a limiting reagent and oxygen gas is an excessive reagent.
Remaining molecules of recessive reagent = 150 - 130 = 20
20 molecules of oxygen gas remains after the reaction.
Moles He = 7.83 x 10^24 / 6.02 x 10^23 =13.0
<span>mass He = 13.0 mol x 4.00 g/mol = 52.0 g</span>
When HCl is added to metal ions, metal chlorides are produced. In this problem, it is asked whether the given ions precipitate or not when added to HCl. According to the rule, all chlorides except Ag+, Pb 2+, Hg2 2+ are soluble. Hence the ion that would precipitate is only lead (II) ion.
<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61
<u>Explanation:</u>
We are given:
Initial moles of iodine gas = 0.100 moles
Initial moles of hydrogen gas = 0.100 moles
Volume of container = 1.00 L
Molarity of the solution is calculated by the equation:



Equilibrium concentration of iodine gas = 0.0210 M
The chemical equation for the reaction of iodine gas and hydrogen gas follows:

<u>Initial:</u> 0.1 0.1
<u>At eqllm:</u> 0.1-x 0.1-x 2x
Evaluating the value of 'x'

The expression of
for above equation follows:
![K_c=\frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
![[HI]_{eq}=2x=(2\times 0.079)=0.158M](https://tex.z-dn.net/?f=%5BHI%5D_%7Beq%7D%3D2x%3D%282%5Ctimes%200.079%29%3D0.158M)
![[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M](https://tex.z-dn.net/?f=%5BH_2%5D_%7Beq%7D%3D%280.1-x%29%3D%280.1-0.079%29%3D0.0210M)
![[I_2]_{eq}=0.0210M](https://tex.z-dn.net/?f=%5BI_2%5D_%7Beq%7D%3D0.0210M)
Putting values in above expression, we get:

Hence, the value of equilibrium constant for the given reaction is 56.61
The mass of the liquid is 280 g.
Mass = 200 mL × (1.4 g/1 mL) = 280 g