All categories

3 years ago

A student isolated 25 g of a compound following a procedure that would theoratically yield 81 g. Wht was his percentt yield?

Chemistry

1 answer:

Sati [7]3 years ago

3 0

<u>Answer:</u> The percent yield of the compound is 30.86 %.

<u>Explanation:</u>

To calculate the percentage yield of a compound, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of compound = 25 g

Theoretical yield of compound = 81 g

Putting values in above equation, we get:

$\%\text{ yield of compound}=\frac{25g}{81g}\times 100\\\\\% \text{yield of compound}=30.86\%$

Hence, the percent yield of the compound is 30.86 %.

You might be interested in

What is the answer??????

balandron [24]

Answer:

Digestion of food.

Explanation:

I hope my answer help you.

5 0

3 years ago

Read 2 more answers

In a reaction of a potential new fuel, it is found that when 2.81 moles of the fuel combusts, 1,612 kJ of energy is released. Wh

Tema [17]

Answer:

-573.67

Explanation:

whenever energy is released in a chemical reaction, we would then expect the delta H of the reaction to be negative because the reaction is an exothermic reaction.

now we have that 2.81 moles of fuel when it combusts would releases 1612kJ of energy

thus, 1 mole will release 1612/2.81 = -573.67kJ of heat

Therefore the delta H of the reaction = -573.67 kJ/mol

3 0

3 years ago

The diagram below show an enlarged view of the beams of a triple-beam balance

VLD [36.1K]

Answer:

1) 455.2 g. 2) 545.2 g.

Explanation:

i think this is right

6 0

3 years ago

What is the percentage yield of O2 if 12.3 g of KClO3 (molar mass 123 g) is decomposed to produce 3.2 g of O2 (molar mass 32 g)

My name is Ann [436]

Answer:

The percentage yield of O2 is 66.7%

Explanation:

Reaction for decomposition of potassium chlorate is:

2KClO₃ → 2KCl + 3O₂

The products are potassium chloride and oxygen.

Let's find out the moles of chlorate.

Mass / Molar mass = Moles

12.3 g / 123 g/mol = 0.1 mol

So ratio is 2:3, 2 moles of chlorate produce 3 mol of oxygen.

Then, 0.1 mol of chlorate may produce (0.1 .3)/ 2 = 0.15 moles

Let's convert the moles of produced oxygen, as to find out the theoretical yield.

0.15 mol . 32 g/ 1mol = 4.8 g

To calculate the percentage yield, the formula is

(Produced Yield / Theoretical yield) . 100 =

(3.2g / 4.8g) . 100 = 66.7 %

8 0

3 years ago

Can someone help me? It needs to have a diagram that has arrows.

daser333 [38]

Answer: The enthalpy change for formation of butane is -125 kJ/mol

Explanation:

The balanced chemical reaction is,

$C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_2O(l)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]$

Putting the values we get :

$\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]$

$-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]$

$H_f_{C_4H_{10}=-125kJ/mol$

Thus enthalpy change for formation of butane is -125 kJ/mol

5 0

3 years ago