Answer : The molarity after a reaction time of 5.00 days is, 0.109 M
Explanation :
The integrated rate law equation for second order reaction follows:
where,
k = rate constant =
t = time taken = 5.00 days
[A] = concentration of substance after time 't' = ?
= Initial concentration = 0.110 M
Now put all the given values in above equation, we get:
Hence, the molarity after a reaction time of 5.00 days is, 0.109 M
Answer:
0.0745 mole of hydrogen gas
Explanation:
Given parameters:
Number of H₂SO₄ = 0.0745 moles
Number of moles of Li = 1.5107 moles
Unknown:
Number of moles of H₂ produced = ?
Solution:
To solve this problem, we have to work from the known specie to the unknown one.
The known specie in this expression is the sulfuric acid, H₂SO₄. We can compare its number of moles with that of the unknown using a balanced chemical equation.
Balanced chemical equation:
2Li + H₂SO₄ → Li₂SO₄ + H₂
From the balanced equation;
Before proceeding, we need to obtain the limiting reagent. This is the reagent whose given proportion is in short supply. It determines the extent of the reaction.
2 mole of Li reacted with 1 mole of H₂SO₄
1.5107 mole of lithium will react with = 0.7554mole of H₂SO₄
But we were given 0.0745 moles,
This suggests that the limiting reagent is the sulfuric acid because it is in short supply;
since 1 mole of sulfuric acid produced 1 mole of hydrogen gas;
0.0745 mole of sulfuric acid will produce 0.0745 mole of hydrogen gas