Which molecule is produced by the transferring of electrons?

The reaction of 9.50 g of carbon with Excess O2 Yield 12.8 of CO2. What is the percent yield of this reaction?

Anna [14]

Answer:

Percentage yield = 36.75%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

C + O₂ —> CO₂

Next, we shall determine the mass of C that reacted and the mass of CO₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of C = 12 g/mol

Mass of C from the balanced equation = 1 × 12 = 13 g

Molar mass of CO₂ = 12 + (2×16)

= 12 + 32

= 44 g/mol

Mass of CO₂ from the balanced equation = 1 × 44 = 44 g

SUMMARY:

From the balanced equation above,

12 g of C reacted to produce 44 g of CO₂.

Next, we shall determine the theoretical yield of CO₂. This can be obtained as follow:

From the balanced equation above,

12 g of C reacted to produce 44 g of CO₂.

Therefore, 9.50 g of C will react to produce = (9.50 × 44) / 12 = 34.83 g of CO₂.

Thus, the theoretical yield of CO₂ is 34.83 g.

Finally, we shall determine the percentage yield of the reaction. This can be obtained as follow:

Actual yield of CO₂ = 12.8

Theoretical yield of CO₂ = 34.83 g

Percentage yield =?

Percentage yield = Actual yield / Theoretical yield × 100

Percentage yield = 12.8 / 34.83 × 100

Percentage yield = 1280 / 34.83

Percentage yield = 36.75%

5 0

3 years ago

A sample of propane (C3H8) has a mass of 0. 47 g. The sample is burned in a bomb calorimeter that has a mass of 1. 350 kg and a

Nady [450]

The amount of heat released by the sample has been 22.54 kJ. Thus, option C is correct.

The specific heat has been defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius.

The specific heat has been expressed as:

$q=mc\Delta T$

<h3 /><h3>Computation for the heat absorbed</h3>

The iron and calorimeter are in side the closed system. Thus, the energy released by the sample, has been equivalent to the energy absorbed by the calorimeter.

$q_{released}=q_{absorbed}\\
q_{released}=m_{calorimeter}\;c_{calorimeter}\;\Delta T$

The given mass of calorimeter has been, $m_{calorimeter}=1350\;\rm g$

The specific heat of the calorimeter has been, $c_{calorimeter}=5.82\;\rm J/g^\circ C$

The change in temperature of the calorimeter has been, $\Delta T=2.87^\circ \rm C$

Substituting the values for heat released:

$q_{released}= 1350\;\text g\;\times\;5.82\;\text J/\text g^\circ \text C\;\times\;2.87^\circ \text C\\
q_{released}=22,549.5\;\text J\\
q_{released}}=22.54\;\rm kJ$