Ask question

Ask question

All categories

2 years ago

12

For the following reaction, 5.20 grams of propane (C3H8) are allowed to react with 22.5 grams of oxygen gas. propane (C3H8) (g)

oxygen (g) carbon dioxide (g) water (g) What is the maximum amount of carbon dioxide that can be formed

1 answer:

Sliva [168]2 years ago

4 0

Answer:

15.58g of CO₂ is the maximum amount that can be produced

Explanation:

The propane reacts with oxygen as follows:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

<em>Where 1 mole of propane reacts with 5 moles of oxygen</em>

To solve this question we need to find the moles of propane and oxygen to find limiting reactant using the chemical reaction:

<em>Moles propane -Molar mass: 44.1g/mol-:</em>

5.20g * (1mol / 44.1g) = 0.118 moles

<em>Moles oxygen -Molar mass: 32g/mol-:</em>

22.5g O₂ * (1mol / 32g) = 0.703 moles

For a complete reaction of 0.703 moles of oxygen are:

0.703 moles O₂ * (1mol C₃H₈ / 5mol O₂) = 0.141 moles of propane are necessaries. As there are just 0.118 moles of propane, <em>propane is limiting reactant.</em>

The moles of carbon dioxide that are produced are:

0.118 moles C₃H₈ * (3 moles CO₂ / 1 mol C₃H₈) = 0.354 moles CO₂

The maximum mass that can be produced is -Molar mass CO₂: 44.01g/mol-:

0.354 moles CO₂ * (44.01g / mol) =

15.58g of CO₂ is the maximum amount that can be produced

You might be interested in

The overlapping of food chains results in a

Katyanochek1 [597]

Answer:

.

Explanation:

8 0

2 years ago

Select the correct answer.

frozen [14]

Answer:

1.79x10^-14

Explanation:

pOH + pH = 14

H+=10^-pH

- Hope that helps! Please let me know if you need further explanation.

3 0

3 years ago

Choose the appropriate electron configuration for an element whose third shell contains six electrons.

slamgirl [31]

For an element whose third shell contains six electrons, the appropriate electron configuration is; 1s2 2s2 2p6 3s2 3p4.

The electron configuration shows the distribution of electrons in the shells of an atom and in orbitals.

We have been told that the six electrons are found in the third shell. This shell has n=3 and the configuration of this shell must ns2 np4.

The only electron configuration that meets this standard is 1s2 2s2 2p6 3s2 3p4.

Learn more: brainly.com/question/18704022

5 0

2 years ago

Pls need help for answer now

yawa3891 [41]

The answer is 0.375 mol of silver chromate

5 0

2 years ago

Never mind jhvjycdtrsesetdfyguhbjnk

OlgaM077 [116]

Complete Question

methanol can be synthesized in the gas phase by the reaction of gas phase carbon monoxide with gas phase hydrogen, a 10.0 L reaction flask contains carbon monoxide gas at 0.461 bar and 22.0 degrees Celsius. 200 mL of hydrogen gas at 7.10 bar and 271 K is introduced. Assuming the reaction goes to completion (100% yield)

what are the partial pressures of each gas at the end of the reaction, once the temperature has returned to 22.0 degrees C express final answer in units of bar

Answer:

The partial pressure of methanol is $P_{CH_3OH_{(g)}} =0.077 \ bar$

The partial pressure of carbon monoxide is $P_{CO} = 0.382 \ bar$

The partial pressure at hydrogen is $P_H = O \ bar$

Explanation:

From the question we are told that

The volume of the flask is $V_f = 10.0 \ L$

The initial pressure of carbon monoxide gas is $P_{CO} = 0.461 \ bar$

The initial temperature of carbon monoxide gas is $T_{CO} = 22.0^oC$

The volume of the hydrogen gas is $V_h = 200 mL = 200 *10^{-3} \ L$

The initial pressure of the hydrogen is $P_H = 7.10 \ bar$

The initial temperature of the hydrogen is $T_H = 271 \ K$

The reaction of carbon monoxide and hydrogen is represented as

$CO_{(g)} + 2H_2_{(g)} \rightarrow CH_3OH_{(g)}$

Generally from the ideal gas equation the initial number of moles of carbon monoxide is

$n_1 = \frac{P_{CO} * V_f }{RT_{CO}}$

Here R is the gas constant with value $R = 0.0821 \ L \cdot atm \cdot mol^{-1} \cdot K$

=> $n_1 = \frac{0.461 * 10 }{0.0821 * (22 + 273)}$

=> $n_1 = 0.19$

Generally from the ideal gas equation the initial number of moles of Hydrogen is

$n_2 = \frac{P_{H} * V_H }{RT_{H}}$

$n_2 = \frac{ 7.10 * 0.2 }{0.0821 * 271 }$

=> $n_2 = 0.064$

Generally from the chemical equation of the reaction we see that

2 moles of hydrogen gas reacts with 1 mole of CO

=> 0.064 moles of hydrogen gas will react with x mole of CO

So

$x = \frac{0.064}{2}$

=> $x = 0.032 \ moles \ of \ CO$

Generally from the chemical equation of the reaction we see that

2 moles of hydrogen gas reacts with 1 mole of $CH_3OH_{(g)}$

=> 0.064 moles of hydrogen gas will react with z mole of $CH_3OH_{(g)}$

So

$z = \frac{0.064}{2}$

=> $z = 0.032 \ moles \ of \ CH_3OH_{(g)}$

From this calculation we see that the limiting reactant is hydrogen

Hence the remaining CO after the reaction is

$n_k = n_1 - x$

=> $n_k = 0.19 - 0.032$

=> $n_k = 0.156$

So at the end of the reaction , the partial pressure for CO is mathematically represented as

$P_{CO} = \frac{n_k * R * T_{CO}}{V}$

=> $P_{CO} = \frac{0.158 * 0.0821 * 295}{10}$

=> $P_{CO} = 0.382 \ bar$

Generally the partial pressure of hydrogen is 0 bar because hydrogen was completely consumed given that it was the limiting reactant

Generally the partial pressure of the methanol is mathematically represented as

$P_{CH_3OH_{(g)}} = \frac{z * R * T_{CO}}{V_f}$

Here $T_{CO}$ is used because it is given the question that the temperature returned to 22.0 degrees C

So

$P_{CH_3OH_{(g)}} = \frac{0.03 * 0.0821 * 295}{10}$

$P_{CH_3OH_{(g)}} =0.077 \ bar$

6 0

2 years ago

Read 2 more answers