If 21.3 grams of lithium react with excess water, how many liters of hydrogen gas can be produced at 297 kelvin and 1.40 atmosph
eres?
1 answer:
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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
Answer:
Both
Explanation:
The combined gas law is also known as the general gas law.
From the ideal gas law we assume that n = 1;
So;
PV = nRT
and then;
=
If we cross multiply;
P₁V₁T₂ = P₂V₂T₁
So;
T₁ =
Also;
V₂ =
So from the choices both are correct