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kherson [118]
2 years ago
7

A certain reaction has a 73.6% yield. if 53.8 grams of the product were predicted by stoichiometry to be made, what would the ac

tual yield be?
Chemistry
1 answer:
krok68 [10]2 years ago
5 0
Actual yield and theoretical yield are usually different.
the percentage yield of product = actual yield / theoretical yield x 100%
the theoretical yield is 53.8 g 
and percentage yield - 73.6 %
from the theoretical yield only 73.6 % of that amount is in the actual yield 
therefore actual yield is 73.6 % of 53.8 g
actual yield - 53.8 x 73.6% = 39.6 g
actual yield is 39.6 g
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Be sure to answer all parts. ΔH o f of hydrogen chloride [HCl(g)] is −92.3 kJ/mol. Given the following data, determine the ident
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Answer:

NH₄Cl(s) → NH₃(g) + HCl(g)

ΔH°rxn 74,89 kJ/mol

Explanation:

The change in enthalpy of formation (ΔHf) is defined as the change in enthalpy in the formation of a substance from its constituent elements. For HCl(g):

<em>(1) </em>¹/₂H₂(g) + ¹/₂ Cl₂ → HCl(g) ΔH = -92,3 kJ/mol

It is possible to sum ΔH of different reactions to obtain ΔH of a global reaction (Hess's law).

For the reactions:

<em>(2) </em>N₂(g) + 4H₂(g) + Cl₂(g) → 2NH₄Cl(s) ΔH°rxn = −630.78 kJ/mol

<em>(3)</em> N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol

The sum of -(2) + (3) gives:

<em>-(2) </em>2NH₄Cl(s) → N₂(g) + 4H₂(g) + Cl₂(g) ΔH°rxn = +630.78 kJ/mol

<em>(3)</em> N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol

<em>-(2) + (3) </em>2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g)

ΔH°rxn = +630.78 kJ/mol −296.4 kJ/mol = +334,38 kJ/mol

Now, the sum of -(2) + (3) + 2×(1)

<em>-(2) + (3) </em>2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g) ΔH°rxn = +334,38 kJ/mol

<em>2×(1)  </em>H₂(g) + Cl₂(g)→ 2HCl(g) ΔH = 2×-92,3 kJ/mol

<em>-(2) + (3) + 2×(1) </em>2NH₄Cl(s) → 2NH₃(g) + 2HCl(g)

ΔH°rxn = +334,38 kJ/mol + 2×-92,3 kJ/mol = 149,78 kJ/mol

The reaction of:

<em>NH₄Cl(s) → NH₃(g) + HCl(g)</em>

<em>Has ΔH°rxn = 149,78kJ/mol / 2 = 74,89 kJ/mol</em>

I hope it helps!

8 0
1 year ago
Methane, CH4, reacts with I2 according to the reaction CH4(g)+I2(g)⇌CH3I(g)+HI(g)
gtnhenbr [62]

Answer:

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

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Kp is the equilibrium constant for the partial pressure of the gases in the reaction, and it is calculated for a general equation:

aA(g) + bB(g) ⇄ cC(g) + dD(g)

Kp = \frac{(pC)^cx(pD)^d}{(pA)^ax(pB)^b}, where p is the partial pressure in the equilibrium. By the reaction given:

CH₄(g) + I₂(g) ⇄ CH₃I(g) + HI(g)

105.1 torr   7.96 torr  0       0            <em> initial partial pressure</em>

-x                  -x            +x     +x          <em> react</em>

105.1-x       7.96-x      x        x            <em>equilibrium</em>

Then:

Kp = \frac{pCH3IxpHI}{pCH4xpI2} = \frac{x^2}{(105.1-x)(7.96-x)}

2.26x10^{-4} = \frac{x^2}{836.596 - 113.06x -x^2}

x² = 0.1891 - 0.0255x -2.26x10⁻⁴x²

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Using Bhaskara's rule:

Δ = (0.0255)² - 4x(0.9997)x(-0.1891)

Δ = 0.7568

x = \frac{-b+/-\sqrt{0.7568} }{2a} = \frac{-0.0255 +/-0.8699}{1.9994}

Using only the positive term, x = 0.42 torr.

So,

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

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