Solve the following stoichiometry problem.

How many atoms of sodium, Na, are required to react completely with 75.0 grams of chlorine using this reaction

2Na + Cl2 --> 2NaCl

OPTION

-6.22 x 1023 atoms of Na

-1.27x1024 atoms of Na

-can't determine with the given information

-1.27 x 1023 atoms of Na

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• How many atoms of sodium, Na, are required to react completely with 75.0 grams of chlorine using this reaction

2Na + Cl2 --> 2NaCl

1. First find the number of moles that chlorine has using the mass we are given.

75 grams of Cl x 70.9 grams/mol = 1.0578 mols of Cl

2. Find the number of mols of Na in the equation by doing a mole-to-mole conversion between Na and Cl

1.0578 mols of Cl x 2 Na/ 1Cl = 2.1156 mols of Na in the reaction.

Now multiply the amount of Na by avagadro's constant (6.02x10^23)

2.1156 mols Na x 6.02x10^23 atoms/1 mol = 1.27x10^24 atoms of sodium

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• 2Na + Cl2 --> 2NaCl

According to the coefficients in this equation, two moles of sodium reacts with one mole of chlorine to produce two moles of sodium chloride. The mass of one mole of Cl2 is 71 grams.

For Cl2, n = 75 ÷ 71

This is approximately 1.06 moles. The number of moles of sodium is twice this number.

For Na, n = 150 ÷ 71

To determine the number of atoms of sodium, multiply the number of moles by 6.02 * 10^23.

N =6.02 * 10^23 * (150 ÷ 71) = 9.03 * 10^25 ÷ 71

The number of moles is approximately 1.27 * 10^24.

OR

For Cl2, N = 6.02 * 10^23 * 75 ÷ 71 = 4.515 * 10^5 ÷ 71

The number of moles of chlorine is approximately 6.36 * 10^23. The number of moles of sodium is twice this number.

For Na, N = 9.03 * 10^25 ÷ 71

This is exactly the same number. Now you know two ways to solve this problem. The second answer is the correct answer. I hope this is helpful for you.

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• learn how to write numbers in exponential format.

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• (75.0 g Cl2) / (70.9064 g Cl2/mol) × (2 mol Na / 1 mol Cl2) × (6.022 × 10^23 atoms/mol) =

1.27 × 10^24 atoms Na

So choose the second option.

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