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saul85 [17]
2 years ago
7

What volume of nitrogen (n2) would be completely consumed in the reaction with 30.80 g of

Chemistry
1 answer:
Shtirlitz [24]2 years ago
5 0

The answer is 285.33g nitrogen would be completely consumed in the reaction with 30.80 g of hydrogen gas.

<h3>What is a mole ?</h3>

A mole is defined as 6.02214076 × 10²³ atoms, molecules, ions, or other chemical units.

Write a balanced equation.

Calculate the moles of H₂ in 30.8 g.

Calculate the moles of N₂ required to react with H₂.

Calculate the mass of N₂.

Calculate the initial mass of N₂.

Start with a balanced equation.

N₂ + 3H₂ --> 2NH₃

Calculate the moles of H₂ in 30.8 g.

n = m/M; where n = moles, m = mass, and M = molar mass.

M(H₂) = 1.008 g/mol

n(H₂) = (30.8 g)/(1.008 g/mol) = 30.56 mol H₂

Calculate the moles of N₂ required to react with 30.56 mol H₂ , using the mole ratio between H₂ and N₂ in the balanced equation.

30.56 mol H₂ × 1 mol N₂/3 mol H₂ = 10.18 mol N₂

Calculate the mass of N₂ in 10.18 mol.

m = n × M

M(N₂) = 2 × 14.007 g/mol N = 28.014 g/mol N₂

m(N₂) = 10.18 mol × 28.014 g/mol = 285.33g N₂

Therefore 285.33g nitrogen would be completely consumed in the reaction with 30.80 g of hydrogen gas.

To know more about mole

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<h3>#a. Answer:</h3>

0.0063 mole

<h3>Solution and explanation:</h3>

We are given 21.0 mL citric acid with a concentration of 0.30 M

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<h3>#b. Answer</h3>

1.21 g citric acid

<h3>Solution</h3>

Part B

We are required to calculate the mass of citric acid in the sample

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Moles of citric acid = 0.0063 mole

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Mass of citric acid = 0.0063 mol × 192.124 g/mol

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<h3>#c. Answer</h3>

4.167 mL

<h3>Solution:</h3>

Part C

We are required to determine the initial volume before dilution;

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Initial concentration (M1) = 0.30 M

Final volume (V2) = 250 mL or 0.25 L

Final concentration (M2) = 0.0050 M

Using the dilution formula we can get the initial volume;

Therefore, since; M1V1 =M2V2

V1 = M2V2÷M1

   = (0.0050 × 0.25)÷ 0.30

   = 0.004167 L or

   = 4.167 mL

Therefore, the initial volume of the solution is 4.167 mL

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