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2 years ago

Using the following bond energies:

Chemistry

1 answer:

uranmaximum [27]2 years ago

4 0

Answer:

The answer is "Option b".

Explanation:

Given equation and value:

$\Rightarrow C_2H_2 \ (g)+ \frac{5}{2}O_2 \ (g) \longrightarrow 2CO_2\ (g)+H_2O\ (g)$

$\left\begin{array}{ccc}C \equiv C&839\\C-H &413\\O=O &495\\ C=O &799\\ O-H &467 \end{array}\right \\\\$

calculating equation value:

$\Rightarrow 839 + 2(413) +\frac{5}{2} (495) \longrightarrow 2(2)799+2 \times 467 +E$

$\Rightarrow 839 + 826 +1237.5 = 3196+934 +E\\\\\Rightarrow 2902.5 =4130 +E\\\\\Rightarrow 2902 = 4130 +E\\\\\Rightarrow 2902 - 4130 +E\\\\\Rightarrow E = -1228 \ KJ \\\\$

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If 125.0g of nitrogen is reacted with 125.0g of hydrogen, what is the theoretical yield of the reaction? What is the excess reac

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Answer:

Hydrogen is the excess reactant

Nitrogen is the limiting reactant

151.6g is theoretical yield

Explanation:

The reaction of N₂ with H₂ to produce NH₃ is:

N₂ + 3H₂ → 2NH₃

To find theoretical yield we need to determine limiting reactant with the moles of each gas as follows:

Nitrogen -Molar mass: 28g/mol-

125.0g * (1mol / 28g) = 4.46 moles

Hydrogen -Molar mass: 2g/mol-

125.0g * (1mol / 2g) = 62.5 moles of hydrogen

For a complete reaction of 4.46 moles of N2 there are needed:

4.46 moles N2 * (3moles H2 / 1mol N2) = 13.38 moles of hydrogen

As there are 62.5 moles of hydrogen:

<h3>Hydrogen is the excess reactant</h3><h3>Nitrogen is the limiting reactant</h3><h3 />

With nitrogen, the limiting reactant, we determine theoretical moles (Assuming 100% of the reaction occurs) and theoretical yield (In mass):

4.46 moles N2 * (2moles NH3 / 1mol N2) = 8.92 moles of ammonia

As molar mass of ammonia is 17g/mol:

8.92 moles of ammonia * (17g/mol) =

<h3>151.6g is theoretical yield</h3>

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Black_prince [1.1K]

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