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andriy [413]
3 years ago
7

The theoretical yield of Cl2 from certain starting amounts of MnO2 and HCl was calculated as 60.25 g and 65.02 g, respectively.

If the percentage yield of Cl2 is 72%, what is its actual yield?
Chemistry
1 answer:
hoa [83]3 years ago
8 0

The actual yield is 43 g Cl₂.

The <em>limiting reactant was MnO₂</em> because it gave the smaller mass of Cl₂.

∴ The <em>theoretical yield</em> is 60.25 g Cl₂.

% yield = actual yield/theoretical yield × 100 %

Actual yield = theoretical yield × (% yield/100 %) = 60.25 g × (72 %/100%) = 43 g

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At a given temperature, 3.12 atm of H2 and 5.52 atm of I2 are mixed and allowed to come to equilibrium. The equilibrium pressure
expeople1 [14]

Answer: 0.596

Explanation:

For this problem, we want to find K_{p}. To do so, we will need to use the ICE chart. The I in ICE is initial quantity. In this case, it is the initial pressure. Pressure is in atm. The C in ICE is change in each quantity. The E is equilibrium.

          H₂(g) + I₂(g) ⇄ 2HI(g)

I          3.12     5.52         0

C      -0.869  -0.869   +1.738

E       2.251    2.251     1.738

<u>For the steps below, refer to the ICE chart above.</u>

1. Since we were given the initial of H₂, I₂ and equilibrium of 2HI, we can fill those into the chart.

2. Since we were not given the initial for 2HI, we will put 0 in their place.

3. For the change, we need to add pressure to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x. We know that the equilibrium of 2HI is 1.738, we know the change is 0.869 because 1.738/2=0.869. Since there are 2 moles of HI, we must divide the equilibrium by 2 to find x, so that we can fill that into the reactants side.

4. With the equilibrium values, we can find the equilibrium pressure. It is products over reactants. We use the values of E.

K_{p} =\frac{[HI]^2}{[H_{2}][I_{2} ] }

The [HI]² comes from 2HI. We more the moles to the exponent when we are calculating the equilibrium pressure.

K_{p} =\frac{(1.738)^2}{(2.251)(2.251)} =0.596

We typically don't put units because K is unitless, but know that the units is atm throughout the entire problem.

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3 years ago
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Answer:

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Explanation:

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3 years ago
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Answer:

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Explanation:

5 0
3 years ago
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What is the empirical formula for a compound that is 29.44\% calcium, 23.55% sulfur, and 47.01% oxygen? This compound is a commo
PSYCHO15rus [73]

Answer:

Empirical formula is CaSO₄.

Explanation:

Given data:

Percentage of calcium =29.44%

Percentage of sulfur = 23.55%

Percentage of oxygen = 47.01%

Empirical formula = ?

Solution:

Number of gram atoms of Ca = 29.44 / 40 = 0.74

Number of gram atoms of S = 23.55 / 32 = 0.74

Number of gram atoms of O = 47.01 / 16 = 3

Atomic ratio:

            Ca                      :        S                :         O

           0.74/0.74           :     0.74/0.74      :       3/0.74

               1                     :          1              :          4

Ca : S : O = 1 : 1 : 4

Empirical formula is CaSO₄.

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3 years ago
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