Asked by Sherry
Check if im right please:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) At a specific time in the reaction, ammonia is disappearing at rate of 0.068 mol/(L • s). What is the corresponding rate of production of water?
I did
Rate of Production H20 = 1/6(0.068)
= 0.0113 mol/(L*s)
Am i right?>
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) At a specific time in the reaction, ammonia is disappearing at rate of 0.068 mol/(L • s). What is the corresponding rate of production of water?
I did
Rate of Production H20 = 1/6(0.068)
= 0.0113 mol/(L*s)
Am i right?>
Answers
Answered by
DrBob222
I would have done it this way/.
0.068 x *6 mols H2O x 4 mols NH3) = 0.068 x 6/4 = ?
0.068 x *6 mols H2O x 4 mols NH3) = 0.068 x 6/4 = ?
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