How many grams of oxygen gas are needed to react completely with 27.442 grams of sulfur dioxide gas? SO2(g) + O2(g) --> SO3(g) (needs to be balanced)

Question

on this problem, please follow the instruction:  to calculate the molar mass, we round the numbers from the periodic table, and usually that is ok. But for problems where you have many sig figs, your molar masses from the periodic table must have at least as many sig figs as your data. Therefore, for each oxygen atom, please use 15.9994 g/mol. For each sulfur atom, please use 32.066 g/mol. This is a 5-sig-fig problem, so when you calculate your molar masses, you must use all of the sig figs available to you from the periodic table. That's how you can get this problem correct. 

Answer 1

First thing to do is balance the equation:

2SO₂(g) + O₂(g) ==> 2SO₃(g) ... balanced equation

Next, we'll get the molar masses for SO₂, and O₂ according to the instructions (5 sig. figs.)

molar mass SO₂ = 32.066 g + 2 * 15.9994 g = 64.065 g / mol

molar mass O₂ = 2 * 15.9994 g = 31.999 g / mole

Now we can use dimensional analysis along with the stoichiometry of the balanced equation to find the grams of O₂ needed to react with 27.442 g of SO₂.

27.442 g SO₂ x 1 mol SO₂ / 64.065 g x 1 mol O₂ / 2 mol SO₂ x 31.999 g O₂ / mol = 6.8533 g O₂ needed