Given the balanced chemical equation below:2 H2 + O2 --> 2 H2OHow many grams of water are produced when 64.0 g O2 gas reacts completely? You will need to calculate molar masses of O2 gas and H2O firstGroup of answer choices2.00 g4.00 g18.02 g72.1 g
Question
Given the balanced chemical equation below:2 H2 + O2 --> 2 H2OHow many grams of water are produced when 64.0 g O2 gas reacts completely? You will need to calculate molar masses of O2 gas and H2O firstGroup of answer choices2.00 g4.00 g18.02 g72.1 g
Solution
First, we need to calculate the molar masses of O2 and H2O.
The molar mass of O2 (Oxygen) is approximately 32.00 g/mol (16.00 g/mol * 2).
The molar mass of H2O (Water) is approximately 18.02 g/mol (1.01 g/mol * 2 for Hydrogen + 16.00 g/mol for Oxygen).
Next, we need to determine the number of moles of O2 in 64.0 g. We do this by dividing the mass of O2 by its molar mass:
64.0 g O2 * (1 mol O2 / 32.00 g O2) = 2.00 mol O2
According to the balanced chemical equation, the ratio of O2 to H2O is 1:2. This means that for every mole of O2 that reacts, 2 moles of H2O are produced.
So, if we have 2.00 mol O2, we will produce 2.00 mol O2 * (2 mol H2O / 1 mol O2) = 4.00 mol H2O
Finally, we convert the moles of H2O to grams using the molar mass of H2O:
4.00 mol H2O * (18.02 g H2O / 1 mol H2O) = 72.1 g H2O
So, 72.1 g of water are produced when 64.0 g O2 gas reacts completely. The correct answer is 72.1 g.
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