What is the mole fraction of o2 in a mixture of 15.1 g of o2, 8.19 g of n2, and 2.47 g of h2?

1 answer:

3 0

The solution is
(15.1 g O2) / (31.99886 g O2/mol) = 0.47189 mol O2 (8.19 g N2) / (28.01344 g N2/mol) = 0.29236 mol N2 (2.48 g H2) / (2.01588 g H2/mol) = 1.2302 mol H2
(0.47189 mol O2) + (0.29236 mol N2) + (1.2302 mol H2) = 1.99445 mol total
(0.47189 mol O2) / (1.99445 mol total) = 0.237 for O2

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Question 2 of 50

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Further Explanation:

To solve this problem, follow the steps below:

Write the balanced chemical equation for the given reaction.
Convert the mass of calcium carbonate into moles.
Determine the number of moles of calcium oxide formed by using the stoichiometric ratio for calcium oxide and calcium carbonate based on the coefficient of the chemical equation.
Convert the number of moles of calcium oxide into mass.

Solving the given problem using the steps above:

STEP 1: The balanced chemical equation for the given reaction is:

$CaCO_{3} \rightarrow \ CaO \ + \ CO_{2}$