Ask question

Ask question

All categories

3 years ago

15

Magnesium oxide can be produced by heating magnesium metal in the presence of oxygen. When 10.1 g of Mg reacts with 10.5 g of O2

, 11.9 g of MgO are collected.
a) Write the balanced equation for this reaction, including all states of matter.
b) What is the limiting reactant in this reaction?
c) What is the theoretical yield for this reaction?
d) What is the percent yield for this reaction?

1 answer:

Alexxandr [17]3 years ago

7 0

Answer:

A. 2Mg(s) + O2(g) —> 2MgO(s)

B. Mg is the limiting reactant.

C. Theoretical yield of MgO is 16.83g.

D. The percentage yield is 70.7%

Explanation:

A. The balanced equation for the reaction. This is given below:

2Mg(s) + O2(g) —> 2MgO(s)

B. Determination of the limiting reactant.

First, we shall determine the mass of Mg and O2 that reacted and the mass of MgO produced from the balanced equation. This is illustrated below:

Molar mass of Mg = 24g/mol

Mass of Mg from the balanced equation = 2 x 24 = 48g.

Molar mass of O2 = 16x2 = 32g/mol.

Mass of O2 from the balanced equation = 1 x 32 = 32g

Molar mass of MgO = 24 + 16 = 40g/mol

Mass of MgO from the balanced equation = 2 x 40 = 80g

Summary:

From the balanced equation above,

48g of Mg reacted with 32g of O2 to produce 80g of MgO.

Now, we can obtain the limiting reactant as follow:

From the balanced equation above,

48g of Mg reacted with 32g of O2.

Therefore, 10.1g of Mg will react with = (10.1 x 32)/48 = 6.73g of O2.

From the calculations made above, we can see that only 6.73g out of 10.5g of O2 given is needed to react completely with 10.1g of Mg.

Therefore, Mg is the limiting reactant and O2 is the excess reactant.

C. Determination of the theoretical yield of MgO.

The limiting reactant is used in this case as it will produce the maximum yield of the reaction since all of I is used up in the reaction.

The theoretical yield can be obtain as illustrated below:

From the balanced equation above,

48g of Mg reacted to produce 80g of MgO.

Therefore, 10.1g of Mg will react to produce = (10.1 x 80)/48 = 16.83g of MgO.

Therefore, the theoretical yield of MgO is 16.83g.

D. Determination of the percentage yield.

This is illustrated below:

Actual yield of MgO = 11.9g

Theoretical yield of MgO = 16.83g

Percentage yield =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 11.9/16.83 x 100

Percentage yield = 70.7%

You might be interested in

Enter your answer in the provided box. When mixed, solutions of barium chloride, BaCl2, and potassium chromate, K2CrO4, form a y

ira [324]

Answer:

35.42g

Explanation:

Step 1:

The balanced equation for the reaction

BaCl2(aq) + K2CrO4(aq) → BaCrO4(s) + 2KCl(aq)

Step 2:

Determination of the limiting reactant.

It is important to determine which of the reactant is limiting the reaction as the limiting reactant is used to determine the maximum yield of the reaction. The limiting reactant can be determined as follow:

From the balanced equation above,

1 mole of BaCl2 reacted with 1 mole of K2CrO4.

Therefore, 0.7 mole of BaCl2 will also react with 0.7 mol of K2CrO4.

From the above illustration, we can see that it requires a higher amount of K2CrO4 to react with 0.7 mol of BaCl2. This simply means that K2CrO4 is the limiting reactant.

Step 3:

Determination of the number of mole of BaCrO4 produced from the reaction.

The limiting reactant is used in this case.

From the balanced equation above,

1 mole of K2CrO4 produced 1 mole of BaCrO4.

Therefore, 0.14 mole of K2CrO4 will also produce 0.14 mole of BaCrO4.

Step 4:

Converting 0.14 mole of BaCrO4 to grams.

This is illustrated below:

Molar Mass of BaCrO4 = 137 + 52 + (16x4) = 137 + 52 + 64 = 253g/mol

Number of mole BaCrO4 = 0.14 mole

Mass of BaCrO4 =?

Mass = number of mole x molar Mass

Mass of BaCrO4 = 0.14 x 253

Mass of BaCrO4 = 35.42g

Therefore, 35.42g of BaCrO4 is produced from the reaction.

4 0

3 years ago

34. The line shows the path of the ball.

kiruha [24]

C. Potential to kinetic

6 0

3 years ago

Read 2 more answers

5. What volume of silver metal will weigh exactly 4500.0 g. The density of silver is 20.5 g/cm.

Vitek1552 [10]

Answer:

The answer is

<h2>219.5 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = \frac{mass}{density} \\$

From the question

mass = 4500 g

density = 20.5 g/cm³

We have

$volume = \frac{4500 }{20.5} \\ = 219.51219...$

We have the final answer as

<h3>219.51 mL</h3>

Hope this helps you

8 0

3 years ago

You have a gas with a density of 4.8 grams per Liter and a molar mass of 75.3 grams per mole. How many liters of gas do you have

Andre45 [30]

d = √((x1 - x2)2 + (y1 - y2)2)

( -2 , 5 ) ( 12 , -1 )

↑ ↑ ↑ ↑

x1 y1 x2 y2

d = √((-2 - 12)2 + (5 - (-1))2) = √((-14)2 + 62) = √(196 + 36) = √232 = 2√58 ≈ 15.23

7 0

3 years ago

Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water

madreJ [45]

Explanation:

The given data is as follows.

Temperature of dry bulb of air = $25^{o}C$

Dew point = $10^{o}C$ = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

Partial pressure of water vapor $(P_{a})$ = vapor pressure of water at a given temperature $(P^{s}_{a})$

Using Antoine's equation we get the following.

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}$

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}$

= 0.17079

$P^{s}_{a}$ = 1.18624 kPa

As total pressure $(P_{t})$ = atmospheric pressure = 760 mm Hg

= 101..325 kPa

The absolute humidity of inlet air = $\frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

$\frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

= 0.00735 kg $H_{2}O$ / kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg $H_{2}O$ / kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

0.07 kg - 0.00735 kg

= 0.06265 kg $H_{2}O$ for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

$0.06265 kg \times 100$

= 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0

3 years ago