A) mixture of 82.49 g Aluminum (26.98 g/mol) and 117.65 g Oxygen (32 g/mol) is allowed to react. What MASS of aluminum oxide (101.96 g/mol) can theoretically be formed.
4AL + 3O2 ---> 2AL2 O3
B) How many grams of NO (30.01 g/mol) are expected if 500g O2 (32 g/mol) reacts completely with excess ammonia (NH3) (17.03 g/mol)?
4NH3 + 5O2 -----> 4NO + 6H2O
PS: I give best answer to those who explain the logic in each step behind the answer.
2007-11-11 12:29:45 · 2 answers · asked by Someguy25 4 in Science & Mathematics ➔ Chemistry
A) 82.5gAl x 1molAl/27gAl = 3.06 moles Al
117.65gO2 x 1molO2/32gO2 = 3.68 moles O2
3.68 moles O2 needs 4/3 x 3.68 = 4.9 mole Al to react, and you don't have that. So O2 is the limiting reagent. It will run out first. So:
3.68molO2 x 2molAl2O3/3molO2 x 102gAl2O3/1molAl2O3 = 250g Al2O3
B) 500gO2 x 1molO2/32gO2 x 4molNO/5molO2 x 30gNO/1molNO = 375gNO
2007-11-11 13:00:28 · answer #1 · answered by steve_geo1 7 · 0⤊ 0⤋
sorry, i'm still in biology.
good luck!! it looks really hard!
2007-11-11 12:38:12 · answer #2 · answered by Anonymous · 0⤊ 0⤋