Calculate the ratio of the effusion ratesbetween chlorine gas (Cl2, molar mass = 71 g/mol) and iodine gas(I2, molar mass = 254 g
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<u>Answer:</u> The amount of water that can be formed is 160 moles
<u>Explanation:</u>
We are given:
Moles of hydrogen gas = 170 moles
Moles of oxygen gas = 80 moles
The chemical equation for the reaction of hydrogen gas and oxygen gas follows:
By Stoichiometry of the reaction:
1 mole of oxygen gas reacts with 2 moles of hydrogen gas
So, 80 moles of oxygen gas will react with = of hydrogen gas
As, given amount of hydrogen gas is more than the required amount. So, it is considered as an excess reagent.
Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of oxygen gas produces 2 moles of water
So, 80 moles of oxygen gas will produce = of water
Hence, the amount of water that can be formed is 160 moles
<u>Answer:</u> The concentration of at equilibrium is 0.00608 M
<u>Explanation:</u>
As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.
In the second dissociation, the ions will remain in equilibrium.
We are given:
Concentration of sulfuric acid = 0.025 M
Equation for the first dissociation of sulfuric acid:
0.025 0.025 0.025
Equation for the second dissociation of sulfuric acid:
<u>Initial:</u> 0.025 0.025
<u>At eqllm:</u> 0.025-x 0.025+x x
The expression of second equilibrium constant equation follows:
We know that:
Putting values in above equation, we get:
Neglecting the negative value of 'x', because concentration cannot be negative.
So, equilibrium concentration of sulfate ion = x = 0.00608 M
Hence, the concentration of at equilibrium is 0.00608 M