Ask question

Ask question

All categories

3 years ago

9

During an investigation, a student burns magnesium to form magnesium oxide. The starting mass of magnesium is measured as 21.3 g

.
2Mg( s)+02( 8) - 2MgO( s)
If the student recovers 30.2 g of magnesium oxide, what is the percent yield? (atomic mass of magnesium = 24.3 amu)

1 answer:

enot [183]3 years ago

3 0

Answer:

Percentage yield = 85.2%

Explanation:

Given data:

Mass of Mg = 21.3 g

Actual yield of MgO = 30.2 g

Percentage yield = ?

Solution:

Chemical equation:

2Mg + O₂ → 2MgO

Number of moles of Mg = mass/molar mass

Number of moles of Mg = 21.3 g / 24.3 g/mol

Number of moles of Mg = 0.88 mol

Now we will compare the moles of MgO with Mg.

Mg : MgO

2 : 2

0.88 : 0.88

Mass of MgO:

Mass of MgO= moles × molar mass

Mass of MgO= 0.88 mol × 40.3g/mol

Mass of MgO = 35.46 g

Actual yield of MgO = 30.2 g

Percentage yield:

Percentage yield = Actual yield/theoretical yield × 100

Percentage yield = 30.2 g/ 35.46 g × 100

Percentage yield = 85.2%

You might be interested in

What is the empirical formula of propene CH3 CH2 CH2

kiruha [24]

Answer:

<u>CH</u>

Explanation:

Molecular formula of propene : <u>C₃H₆</u>

Take the HCF of carbon and hydrogen atoms :

3 = 3
6 = 2 x 3

Then, we can write the formula as :

3CH
This means there are 3 moles present

Empirical Formula :

Molecular Formula / No. of moles
C₃H₆ / 3
<u>CH</u>

<u></u>

The empirical formula of propene is <u>CH</u>

6 0

2 years ago

What was the weight percent of water in the hydrate before heating?

DedPeter [7]

Data:

weight of water before heating = 0.349

weight of hydrate before heateing = 2.107

Formula:

Weight percent of water = [ (weight of water) / (weight of the hydrate) ] * 100

Solution:

Weight percent of water = [ 0.349 / 2.107] * 100 ≈ 16.6 %

Answer: 16.6%

3 0

3 years ago

If liquid water is exposed to normal atmospheric pressure, what needs to change in order to change its state of matter?

inn [45]

Answer: Option (b) is the correct answer.

Explanation:

State of a substance changes when heat is provided to a substance.

This is because when we heat water then intermoleclar forces present within its molecules tend to break down. Due to this molecules start to move away from each other.

As a result, kinetic energy of molecules increases and they collide rapidly. Hence, solid state of water changes into liquid state and upon excessive heating liquid state of water changes into vapor state.

Thus, we conclude that temperature of water needs to change in order to change its state of matter.

6 0

3 years ago

Read 2 more answers

How many electrons in the 4 and 5th shell?

patriot [66]

The fifth shell can hold up to 50 electrons and the forth shell can hold 8 electrons

8 0

3 years ago

Substance ΔG°f(kJ/mol) M3O4(s) −8.80 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represe

baherus [9]

Answer:

The equilibrium constant is $K =0.02867$

The equilibrium pressure for oxygen gas is $P_{O_2} = 0.09367\ atm$

Explanation:

From the question we are told that

The equation of the chemical reaction is

$M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}$

The Gibbs free energy for $M_3 O_4_{(s)}$ is $\Delta G^o_{1} = -8.80 \ kJ/mol$

The Gibbs free energy for $M{(s)}$ is $\Delta G^o_{2} = 0 \ kJ/mol$

The Gibbs free energy for $O_2{(s)}$ is $\Delta G^o_{3} = 0 \ kJ/mol$

The Gibbs free energy of the reaction is mathematically represented as

$\Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r$

$\Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)$

Substituting values

From the balanced equation

$\Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]$

$\Delta G^o_{re} = 8.80 kJ/mol =8800J/mol$

The Gibbs free energy of the reaction can also be represented mathematically as

$\Delta G^o_{re} = -RTln K$

Where R is the gas constant with a value of $R = 8.314 J/mol \cdot K$

T is the temperature with a given value of $T = 298 K$

K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction is mathematically represented as

$K_p =[ P_{O_2}]^{\frac{3}{2} }$

Where $[ P_{O_2}]$ is the equilibrium pressure of oxygen

The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the second equation of the Gibbs free energy of the reaction

$K = e^{- \frac{\Delta G^o_{re}}{RT} }$

Substituting values

$K= e^{\frac{8800}{8.314 * 298} }$

$K =0.02867$

Now substituting this into the equation above to obtain the equilibrium of oxygen

$0.02867 = [P_{O_2}]^{[\frac{3}{2} ]}$

multiplying through by $1 ^{\frac{2}{3} }$

$P_{O_2} = [0.02867]^{\frac{2}{3} }$

$P_{O_2} = 0.09367\ atm$

3 0

3 years ago