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faltersainse [42]
1 year ago
5

What is the percent yield of this reaction if 22.o g of Mgl2 is produced by the reaction of 25.0 g of Mg with 25.0 g of l2?

Chemistry
1 answer:
expeople1 [14]1 year ago
8 0

% yield = 80.719

<h3>Further explanation</h3>

Given

22.0 g of Mgl₂

25.0 g of Mg

25.0 g of l₂

Required

The percent yield

Solution

Reaction

Mg + I₂⇒ MgI₂

mol Mg = 25 g : 24.305 g/mol = 1.029

mol I₂ = 25 g : 253.809 g/mol = 0.098

Limiting reactant = I₂

Excess reactant = Mg

mol MgI₂ based on I₂, so mol MgI₂ = 0.098

Mass MgI₂ (theoretical):

= mol x MW

= 0.098 x 278.114

= 27.255 g

% yield = (actual/theoretical) x 100%

% yield = (22 / 27.255) x 100%

% yield = 80.719

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mr_godi [17]

Answer: Thus the yield of uranium from 2.50 kg U_3O_8 is 2.12 kg

Explanation:

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To calculate the number of moles, we use the equation:

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As 1 mole of U_3O_8 contains = 3 moles of U

2.97 mole of U_3O_8 contains = \frac{3}{1}\times 2.97=8.91moles moles of U

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8 0
1 year ago
At –45oC, 71 g of fluorine gas take up 6843 mL of space. What is the pressure of the gas, in kPa?
Anika [276]
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4 0
2 years ago
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5 0
1 year ago
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Firdavs [7]
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4 0
2 years ago
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4 0
2 years ago
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