Question

In the reaction represented above, what mass of HF is produced by the reaction of 3.0 x 10^23 molecules of H_2 with excess F2? (Assume the reaction goes to completion.)

a) 1.5 g
b) 3.0 g
c) 6.0 g
d) 9.0 g

Answer 1

The calculation yields 10.005 g of HF produced from 3.0 x 10^23 molecules of H2, but this option is not listed in the given choices. Therefore, there seems to be a discrepancy either in the provided number of molecules or in the answer choices presented.

Step-by-step explanation:

To determine the mass of HF produced from 3.0 x 1023 molecules of H2, we use Avogadro's number which is 6.02 x 1023 molecules per mole, as a conversion factor. First, we find the number of moles of H2 by dividing the number of molecules by Avogadro's number:

• 3.0 x 1023 molecules H2 / 6.02 x 1023 molecules/mol

This gives us 0.5 moles of H2. The reaction between H2 and F2 to produce HF is:

• H2 + F2 → 2HF

Since according to the balanced equation, one mole of H2 produces two moles of HF, 0.5 moles of H2 would produce 1 mole of HF. The molar mass of HF is approximately 20.01 g/mol. Therefore, the mass of HF produced is:

• 1 mole of HF * 20.01 g/mol = 20.01 g

But since we only have half a mole of H2, we must divide the mass of HF by 2:

• 20.01 g / 2 = 10.005 g

However, since the question asks for an answer with three significant figures and the options provided do not contain 10.005 g, we recognize a mistake in our calculation or in the question's provided options. Either the number of molecules provided in the question is incorrect, or there may be an error in the answer choices.