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ira [324]
3 years ago
8

A student collects 350 mL of a vapor at a temperature of 67°C. The atmospheric pressure at the time of collection is 0.900 atm.

How many moles of the gas does the student collect?
0.011 moles.
0.053 moles
0.11 moles
0.53 moles
Chemistry
1 answer:
castortr0y [4]3 years ago
8 0

Answer: First option 0.011 moles

Explanation:

You need to use the formula for Ideal gases which is:

PV = nRT

Where:

P = Pressure (in atm)

V = Volume (In Liters)

n = moles

R = gas constant (this depends of the units; in this case is 0.082 L atm/K)

T = temperature (In K)

From the formula above, we can solve for n:

n = PV / RT

Let's convert first the temperature and volume to K and L respectively:

T = 67 + 273 = 340 K

V = 350 mL / 1000 = 0.35 L

Finally, let's put all the values in the formula above to solve for the value of n:

n = 0.9 * 0.35 / 0.082 * 340

n = 0.315 / 27.88

n = 0.011 moles

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Ammonia and oxygen react to form nitrogen monoxide and water. Construct your own balanced equation to determine the amount of NO
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NO would form 65.7 g.

H₂O would form 59.13 g.

Explanation:

Given data:

Moles of NH₃ = 2.19

Moles of O₂ = 4.93

Mass of NO produced = ?

Mass of  produced H₂O = ?

Solution:

First of all we will write the balance chemical equation,

4NH₃ + 5O₂   →   4NO + 6H₂O

Now we will compare the moles of NO and H₂O with ammonia from balanced chemical equation:

NH₃  :   NO                                   NH₃  :   H₂O

4     :    4                                          4    :      6

2.19   :    2.19                                 2.19  : 6/4 × 2.19 = 3.285 mol

Now we will compare the moles of NO and H₂O with oxygen from balanced chemical equation:

O₂  :   NO                                               O₂ :   H₂O

5     :    4                                                  5     :    6

4.93   :   4/5×4.93 = 3.944 mol               4.93  : 6/5 × 4.93 = 5.916 mol

we can see that moles of water and nitrogen monoxide produced from the ammonia are less, so ammonia will be limiting reactant and will limit the product yield.

Mass of water = number of  moles × molar mass

Mass of water = 3.285 mol × 18 g/mol

Mass of water = 59.13 g

Mass of nitrogen monoxide  = number of  moles × molar mass

Mass of nitrogen monoxide = 2.19 mol × 30 g/mol

Mass of nitrogen monoxide = 65.7 g

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