If 65.8 grams of SO3 are produced in the unbalanced reaction below, how many grams of oxygen reacted? SO2(g) + O2(g) —> SO3(g) Show work please.

13.13g

Explanation:

65.8÷80.06 =0.821 mol the same

the same amount of moles of Oxygen reacts with

because the ratio is 1 :1-1

let's calculate how much grams for 0.821 of oxygen

oxygen 1 mole is =16 g

well for 0.821=16* 0.821 which is 13.136grams

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

the answer to your question is c