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2 years ago

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]2 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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A lead–tin alloy of composition 30 wt% Sn–70 wt% Pb is slowly heated from a temperature of 150°C (300°F).(a) At what temperature

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Answer:

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So the first liquid phase forms at 231.9 °C and because Pb starts melting at a higher temperature, that phase's composition will be 100% Sn.

The mixture will be completely melted when you are a the higher melting temperature of all components (in this case Pb), so it will all in liquid phase at 327.5 °C.

At that temperature all Sn was already in liquid state and, therefore, the last solid's composition will be 100% Pb.

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Explanation:

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$\lambda = \frac{h}{p}$

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p = momentum = $mass \times velocity$

Putting the values in the formula as follows.

$\lambda = \frac{h}{mass \times velocity}$

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= $2.037 \times 10^{-34} m$

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