Consider the reaction 2H2 + O2 --> 2H2O
What is the mass of water produced when 11.0 grams of O2 reacts with excess H2?
Is the answer 22.0 grams of H2O? That's what I got
No it isn't 22.0 g but I think I know what you did wrong. I think you said that 11.0 g O2 x 2 for the coefficient for H2O gives 22.0 grams. It works that way EXCEPT not with grams. The equations work with moles and not grams.
So mols O2 = g O2/molar mass O2 = 11.0 g/32 = 0.344
Now convert to mols H2O produced this way.
0.344 mols O2 x (2 mols H2O produced/1 mol O2) = 0.344 x 2 = 0.688
Then grams H2O = mols H2O x molar mass H2O = 0.688 mols x 18 g/mol = ? grams H2O
To determine the mass of water produced in the reaction, you need to use stoichiometry. Stoichiometry is a calculation of the quantitative relationships between reactants and products in a chemical reaction.
First, you need to calculate the number of moles of O2. To do this, divide the given mass by the molar mass of O2 (which is 32.00 g/mol).
11.0 g O2 / 32.00 g/mol O2 = 0.34375 mol O2
According to the balanced equation, for every 1 mole of O2, 2 moles of H2O are produced. So, you can conclude that 0.34375 mol O2 will produce (2 moles of H2O / 1 mole of O2) * 0.34375 mol O2 = 0.6875 mol H2O.
Finally, to find the mass of water, multiply the number of moles of H2O by its molar mass (which is 18.02 g/mol).
0.6875 mol H2O * 18.02 g/mol H2O = 12.34 g H2O
Therefore, the correct mass of water produced when 11.0 grams of O2 react with excess H2 is 12.34 grams of H2O, not 22.0 grams.