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mixer [17]
3 years ago
11

You have 2.2 mol Xe and 2.0 mol F₂, but when you carry out the reaction you end up with only 0.25 mol XeF₄. What is the percent

yield of this experiment? Xe(g) + 2 F₂ (g) → XeF₄ (g)
Chemistry
1 answer:
alexira [117]3 years ago
8 0

Answer:

The correct answer is 25 %

Explanation:

According to the chemical reaction:

Xe(g) + 2 F₂ (g) → XeF₄ (g)

1 mol of Xe(g) reacts with 2 mol of F₂(g), so the stoichiometric ratio os reactants is 2 mol F₂/mol Xe.

We have 2.2 mol Xe and 2.0 mol F₂, so the ratio is:

2.0 mol F₂/2.2 mol Xe = 0.909 mol F₂/mol Xe

The molar ratio of reactant we have is lower than the required, so F₂ is the limiting reactant.

By using the limiting reactant, we calculate the theoretical amount of product (XeF₄). For this, we know that 1 mol of XeF₄ is formed from 2 mol of F₂ (1 mol XeF₄/ 2 mol F₂), and we have 2.0 mol F₂:

2.0 mol F₂ x (1 mol XeF₄/ 2 mol F₂)= 1 mol XeF₄

If we only obtained 0.25 mol XeF₄, the percent yield of the experiment is:

Yield = experimental amount/theoretical amount x 100%

        = 0.25 mol XeF₄/ 1 mol XeF₄ x 100% = 25 %

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What mass of butane in grams is necessary to produce 1.5×103 kj of heat what mass of co2 is produced?
kari74 [83]
The heat of reaction (i.e. combustion) of butane (C_{4} H_{10}) when reacted with oxygen (O_{2})  is -2658 kJ/mol butane, and the chemical reaction is given by: 

C_{4} H_{10} + \frac{13}{2} O_{2} ---> 4 CO_{2}  + 5 H_{2}O

The mass of butane required in the reaction is based on the heat produced by the reaction, which is given to be -1,500 kJ. The minus sign is added because the reaction releases heat (exothermic), which means that the products are in a "lower energy state" than the reactants. 

Dividing this with the heat of reaction per mole of butane reacted would give the number of moles butane required. Then, multiplying the answer with the molar mass of butane which is 58 grams/mole, will give the mass of butane required. 

Moles of butane = [(-1,500 kJ)/(-2658 kJ/mol butane)]
Moles of butane = 0.5643 moles butane

Mass of butane  = 0.5643 moles butane * 58 grams/mol butane
Mass of butane  = 32.73 grams butane

The mass of carbon dioxide (CO_{2}) can be determined by multiplying the moles of butane (C_{4} H_{10}) with the mole ratio of (CO_{2}) produced to the (C_{4} H_{10}) reacted, and then with the molar mass of (CO_{2}), which is 44 grams/mole. 

Mass of carbon dioxide produced 
    = 0.5643 moles butane * [4 moles CO_{2}/ 1 mole C_{4} H_{10}] * 44 grams/mole CO_{2}

Mass of carbon dioxide produced  
    = 99.32 grams CO_{2}

Thus, the mass of butane required is 32.73 grams, and the mass of carbon dioxide produced from the reaction of this amount of butane is 99.32 grams. 
                
4 0
3 years ago
Read 2 more answers
Need some help with this problem please!
Tom [10]

Answer:

18 liters

Explanation:

Step 1: Figure out what the formula and what you are dealing with.

- 25 degrees celcius is constant, so it is irrelevant for the mathmatical part.

- P1 = 1 atm

- P2 = 20 atm

- V1 = 360 liters

- V2 = trying to find

Note: remember the original equation is V1/P1 = V2/P2

- Step 2: Rearrange the equation to fit this problem, you should get...

V2 = V1 x P1 / P2

- Step 3: Fill our own numbers in. You should get...

360 L x 1 atm / 20 atm = 18 Liters (do the math)

- Answer = 18 Liters

- Remember to just follow the formula and fill it in with your own numbers.

If you need any more help comment below. I am happy to help anytime.

7 0
3 years ago
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