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Anestetic [448]
2 years ago
5

If 68.0 grams of n2, are able to completely react with unlimited hydrogen, how many grams of ammonia nh3, will be produced

Chemistry
1 answer:
lions [1.4K]2 years ago
3 0
N2 +3 H2 ⇌ 2 NH3
no of mol of N2 used
=68/28
=2.429 mol
1 mol of N2 produces 2 mol of NH3
2.429 mol of N2 produces 4.858 mol of NH3
mass of NH3 produced
=4.858×17
=82.586g
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How many grams of water are produced when 4.50 L of
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The answer for the following problem has been mentioned below.

  • <em><u>Therefore the mass of the water is 5.802 grams.</u></em>

Explanation:

Given:

volume of oxygen (V) = 4.50 L

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We know;

According to the ideal gas equation;

     P × V = n × R × T

As we know;

no of moles = \frac{m}{M}

m represents the mass of oxygen (m)

M represents the Gram molecular mass (M)

According to above mentioned equation;

           P × V = n × R × T

P represents the pressure of the oxygen

V represents the volume of the oxygen

n represents the no of moles of the oxygen

R represents the universal gas constant

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Substituting the values in the above equation;

                  2.50 × 4.50 = \frac{m}{16.0} × 0.0821 × 425

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As we know;

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m_{1} represents the mass of the oxygen

M_{1} represents the gram molecular mass of the oxygen

m_{2} represents the mass of the water

M_{2} represents the gram molecular mass of water

    From the above given formula,

      \frac{5.158}{16.0} = \frac{m_{2} }{18}

where;

Gram molecular weight of water = 18.0 u

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<em><u>Therefore the mass of the water is 5.802 grams.</u></em>

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