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3 years ago

Be sure to answer all parts. How many moles of O are needed to combine with 0.276 mole of C to form CO

Chemistry

1 answer:

Vikki [24]3 years ago

7 0

Answer:

0.138 moles of oxygen are needed to combine with 0.276 moles of carbon.

Explanation:

Given data:

Moles of oxygen needed = ?

Moles of carbon present = 0.276 mol

Solution:

Chemical equation:

2C + O₂ → 2CO

Now we will compare the moles of carbon with oxygen.

C : O₂

2 : 1

0.276 : 1/2×0.276 = 0.138 mol

Thus, 0.138 moles of oxygen are needed to combine with 0.276 moles of carbon.

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3 years ago

Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)T

Eduardwww [97]

Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Explanation:

(a)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 730.5 mmHg

V = Volume of the gas = 9.49 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$730.5mmHg\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 293}=0.3794mol$

According to the reaction shown below as:-

$2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)$

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when $\frac{2}{3}\times 0.3794$ moles of potassium chlorate undergoes reaction.

Moles of potassium chlorate reacted = 0.2529 moles

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 735.5 mmHg

V = Volume of the gas = 9.99 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$735.5mmHg\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 293}=0.40211mol$

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol\times 32g/mol)=12.8675g$

Hence, the amount of oxygen gas collected will be 12.8675 g

5 0

3 years ago

I bet no one can solve this

Nikitich [7]

The answer is 89.4 g/mol cuz the two units carry positives so that means 8.50 flows for 2 more hours than usual.

3 0

3 years ago

15. A pipe leaks water at a rate of 1.24 mL/s. What is the rate of the water leak in L/hr? ALL

alex41 [277]

Answer:

4.46

Explanation:

1.24 ml= 0.00124 L

3600 s=1 hr

0.00124 x 3600=4.46

8 0

3 years ago