Given the following equation: 2h20=2h2+02 how many grams of o2 are produced if 3.6 grams of h20 react?
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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
Moles of potassium permanganate = 0.0008
<h3>Further explanation </h3>Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution). Determination of the endpoint/equivalence point of the reaction can use indicators according to the appropriate pH range
Reaction
5Na2C2O4(aq) + 2KMnO4(aq) + 8H2SO4(aq) ---> 2MnSO4(aq) + K2SO4(aq) + 5Na2SO4(aq) + 10CO2(g) + 8H2O(1)
The end point ⇒titrant and analyte moles equal
titrant : potassium permanganate-KMnO4
analyte : sodium oxalate - Na2C2O4
so moles of KMnO4 = moles of Na2C2O4
moles of Na2C2O4(mass = 0.2640 g, MW=134 g/mol) :
From equation, mol ratio Na2C2O4 : KMnO4 = 5 : 2, so mol KMnO4 :