posted by Anonymous .
When hydrogen selenide gas reacts with oxygen, selenium dioxide and steam are produced.
How many liters of selenium dioxide would be produced from 5.50 L of oxygen? Assume 90% yield and that all gases are measured at the same temperature and pressure.
Balance the equation:
2H2Se + 3O2 > 2SeO2 + 2H2O (g)
For every three volumes of oxygen, on gets two volumes of SeO2. Here, .90 yield.
Volume SeO2= .90*2/3 * 5.5
check my thinking.
I am a little disturbed that we are calculating SeO2 as a gas. If SiO2 is a solid, the would we not expect SeO2 to be a solid? My Merck Index lists it as a solid. At any rate, the other post is correct but here is how you do it in mols.
2H2Se + 3O2 ==> 2SeO2 + 2H2O(g)
mols oxygen = 5.5L/22.4 = 0.2455.
mols SeO2 produced = 0.2455 mols O2 x (2 mols SeO2/2 mols O2) = 0.164 mols SeO2.
vol SeO2 = 0.164 mol x 22.4 L/mol =3.67 L.
The yield is only 90%; therefore, we would obtain 3.67 x 0.9 = 3.3 L
I hope this helps.