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Over [174]
3 years ago
6

Sulfuric acid, H 2 S O 4 , is an important industrial chemical, typically synthesized in a multi-step process. What is the perce

nt yield if a batch of H 2 SO 4 has a theoretical yield of 3.9 kg, and 2.8 kg are obtained at the end of the process
Chemistry
1 answer:
STatiana [176]3 years ago
3 0

Answer:

71.8%

Explanation:

The percent yield of a process or a chemical reaction is calculated as:

percent yield = actual yield/theoretical yield x 100%

From the problem we have:

theoretical yield = 3.9 kg

actual yield = 2.8 kg

Thus, we calculate the yield of H₂SO₄ production as follows:

percent yield = (2.8 kg/3.9 kg) x 100 = 71.79 % ≅ 71.8%

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Why are covalent bonds between hydrogen and nitrogen or oxygen polar?
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The answer to your question is below

Explanation:

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Read 2 more answers
How many molecules of XeF6 are formed from 12.9 L of F2 (at 298 K and 2.6 atm) according to 11) the following reaction? Assume t
ddd [48]

Answer:

#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.

Explanation:

Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆

1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.

=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP

2. Calculate moles of F₂ used

=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used

3. Calculate moles of XeF₆ produced from reaction ratios …

Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆

4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number  (6.02 x 10²³ molecules/mole)

=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)

                                  = 2.75 x 10²³ molecules XeF₆.

8 0
3 years ago
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