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weeeeeb [17]
3 years ago
13

10.0 g of Mg are reacted with 95.0 g of I2. If 27.5 g of magnesium iodide are obtained, what is the percent yield

Chemistry
1 answer:
Tom [10]3 years ago
8 0

Answer:

% yield of reaction is 26.4

Explanation:

The reaction is:

Mg + I₂ →  MgI₂

Our reactants are magnessium and iodine. We determine the moles of each to find the limiting reactant:

10 g . 1mol / 24.3 g = 0.411 moles of Mg

95 g . 1mol / 253.8g = 0.374 moles of I₂

Ratio is 1:1. For 1 mol of Mg we need 1 mol of iodine

For 0.411 moles, we need the same amount, but we only have 0.374 moles of iodine, that's why the gas is the limiting reactant.

As ratio is 1:1 again, 0.374 moles of iodine can produce 0.374 moles of MgI₂

We determine the mas (theoretical yield): 0.374 mol . 278.1 g/mol = 104 g

To calculate the percent yield:

% yield = (yield produced /theoretical yield) . 100

% yield = (27.5 g/ 104g) . 100 = 26.4

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Answer:

B) -4.1 units

Explanation:

According to this question, a state property X has a value 89.6 units. It undergoes the certain changes as follows:

- first increase by 3.6 units

- then increase by another 18.7 units

- then decrease by 12.2 units

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This can be mathematically represented by 89.6 - {3.6 + 18.7 - 12.2 - x) = 85.5

To get x, we say;

89.6 + 3.6 = 93.2

93.2 + 18.7 = 111.9

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The changes that occured is represented as follows:

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3 years ago
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Reference link for the Chloroacetate structure.

https://www.google.com/search?q=lewis+structure+for+CH2ClCOO-&prmd=ivn&sxsrf=ALeKk03mQcLiY-q5pEriMR0_26ZTXLjmJg:1589680325594&source=lnms&tbm=isch&sa=X&ved=2ahUKEwjfxPPY5LnpAhVloXEKHeAwD-wQ_AUoAXoECA4QAQ&cshid=1589680746615&biw=360&bih=559&dpr=3#

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MissTica

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