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2 years ago

You start with 100mL of 14M HNO3 and 10grams of copper metal. How many moles of the nitrogen dioxide gas will be produced?

Chemistry

1 answer:

givi [52]2 years ago

7 0

Answer:

0.316 moles are produced.

Explanation:

We state the redox reaction:

4HNO₃ + Cu → Cu(NO₃)₂ + 2NO₂ + 2H₂O

We need to determine the limting reactant:

0.1 L . 14 M = 1.4 moles

10 g . 1mol/ 63.54g =0.158 mol

Cu is the limiting reactant. Let's see

4 moles of acid need 1 mol of Cu to react

1.4 moles of acid may react to (4 . 1) / 1.4 = 0.35 moles

We do not have enough Cu.

1 mol of Cu can produce 1 mol of NO₂

Then 0.158 moles will produce, 0.316 moles. (double of moles)

If we see stoichiometry, ratio is 1:2

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Since you have not included the chemical reaction I will explain you in detail.

1) To determine the limiting agent you need two things:

- the balanced chemical equation

- the amount of every reactant involved as per the chemical equation

2) The work is:

- state the mole ratios of all the reactants: these are the ratios of the coefficientes of the reactans in the balanced chemical equation.

- determine the number of moles of each reactant with this formula:

number of moles = (mass in grams) / (molar mass)

- set the proportion with the two ratios (theoretical moles and actual moles)

- compare which reactant is below than the stated by the theoretical ratio.

3) Example: determine the limiting agent in this reaction if there are 100 grams of each reactant:

i) Chemical equation: H₂ + O₂ → H₂O

ii) Balanced chemical equation: 2H₂ + O₂ → 2H₂O

iii) Theoretical mole ration of the reactants: 2 moles H₂ : 1 mol O₂

iv) Covert 100 g of H₂ into number of moles

n = 100g / 2g/mol = 50 mol of H₂

v) Convert 100 g of O₂ to moles:

n = 100 g / 32 g/mol = 3.125 mol

vi) Actual ratio: 50 mol H₂ / 3.125 mol O₂

vii) Compare the two ratios:

2 mol H₂ / 1 mol O ₂ < 50 mol H₂ / 3.125 mol O₂

Conclusion: the actual ratio of H₂ to O₂ is greater than the theoretical ratio, meaning that the H₂ is in excess respect to the O₂. And that means that O₂ will be consumed completely while some H₂ will remain without react.

Therefore, the O₂ is the limiting reactant in this example.

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