How many grams of oxygen gas are needed to react completely with 28.442 grams of sulfur dioxide gas ?

SO2(g) + O2(g) SO3(g) (needs to be balanced. Balance it by placing a 1/2 in front of the 02.) 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. Melissa Maribel likes to 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. Thank you.

jsimpson11000 jsimpson11000 · Answer 2 · 2022-07-29T15:01:31+02:00

Answer:

7.1029 gms of O2

Explanation:

First balance the equation to get

2 SO2 + O2 ====> 2 SO3

Mole weight of sulfur dioxide gas

S + 2 O == 32.066 + 2 * 15.999 = 64.064 gm/mole ( 5 signif digits)

28.442 gm / 64.064 gm/mole = .44396 mole of SO2 ( 5 sig digits)

From the balanced equation, you can see you need 1/2 as many moles of O2 as SO2:

mole weight of O2 = 2 * 15.999 = 31.998 gm / mole ( 5 sig digits)

now you need 1/2 of .44396 moles of this

1/2 * .44396 * 31.998 = 7.1029 gm ( 5 sig digits)