Define in your own words: Solute, Solvent and Solution.
1 answer:
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<u>Answer:</u> The theoretical yield of iron(III) sulfate is 26.6 grams
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of iron(III) phosphate = 20.00 g
Molar mass of iron(III) phosphate = 150.82 g/mol
Putting values in equation 1, we get:
The given chemical equation follows:
As, sodium sulfate is present in excess. So, it is considered as an excess reagent.
Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate
So, 0.133 moles of iron(III) phosphate will produce = of iron(III) sulfate
Now, calculating the mass of iron(III) sulfate from equation 1, we get:
Molar mass of iron(III) sulfate = 399.9 g/mol
Moles of iron(III) sulfate = 0.0665 moles
Putting values in equation 1, we get:
Hence, the theoretical yield of iron(III) sulfate is 26.6 grams
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The volume of H₂O = 5 L
<h3>Further explanation</h3>Given
5L of H₂ and 3L O₂
Reaction
2H₂ (g) + O₂(g) ⇒2H₂O(g)
Required
The volume of H₂O
Solution
Avogadro's hypothesis:
<em>In the same T,P and V, the gas contains the same number of molecules </em>
So the ratio of gas volume will be equal to the ratio of gas moles
mol H₂ = 5, mol O₂ = 3
- Find limiting reactants
From equation, mol ratio H₂ : O₂ = 2 : 1, so :
- Find volume H₂O
mol H₂O based on mol H₂, and from equation mol ratio H₂ : H₂O=2 : 2, so mol H₂O = 5 mol and the volume also 5 L