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juin [17]
2 years ago
9

How many grams of KOH are required to make 125mL of a .500M solution?

Chemistry
1 answer:
Bumek [7]2 years ago
3 0

Answer:

Answer:3.51 g

Explanation:

M(KOH)= 39.10 + 16 + 1.008 = 56.108 g/molAmount of substance is n= m/M = cV and mass m= McVolume in litres is 0.125 l

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Hydrochloric acid has the chemical formula HCl. A certain quantity of hydrochloric acid has 2.39x1024 atoms. How many moles of h
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What mass of Al2(SO4)3 results from mixing 12.93 g of AlCl3 with 10.34 g of (NH4)2SO3 as shown with a 54.2% yield?
avanturin [10]

Mass of Al₂(SO₄)₃ : 4.822 g

<h3>Further explanation  </h3>

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.  

Reaction

2AlCl₃ + 3(NH₄)₂SO₄⇒Al₂(SO₄)₃+ 6NH₄Cl

MW AlCl₃ :133,34 g/mol

MW (NH₄)₂SO₄ : 132,14 g/mol

MW Al₂(SO₄)₃ : 342,15 g/mol

mol  AlCl₃

\tt mol=\dfrac{12.93}{133.34}=0.097

mol (NH₄)₂SO₄

\tt mol=\dfrac{10.34}{132.14}=0.078

Limitng reactants (ratio mol : coefficient = the smaller)

AlCl₃ : (NH₄)₂SO₄ =

\tt \dfrac{0.097}{2}:\dfrac{0.078}{3}=0.0485:0.026

(NH₄)₂SO₄ ⇒ limiting reactants

So mol Al₂(SO₄)₃ from (NH₄)₂SO₄

\tt \dfrac{1}{3}\times 0.078=0.026

mass Al₂(SO₄)₃

\tt 0.026\times 342.15=8.896

with 54.2% yield, the mass of Al₂(SO₄)₃

\tt 0.542\times 8.896=4.822~g

7 0
3 years ago
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