Determine the formula of the oxide.
0.771/52 = about 0.015
0.712/16 = 0.045
CrO3 is the compound.
Cr + O2 ==> CrO3.
You balance.
In an experiment, 0.771g Cr(s)reacts completely with 0.712g O2 to form a single solid compound.
Write a balanced equation
3 answers
I balanced the equation and got
2Cr+3O2->2CrO3.. but I was informed that this is the incorrect answer
2Cr+3O2->2CrO3.. but I was informed that this is the incorrect answer
I've thought about this for the last two hours and I don't think this is wrong.
0.771g Cr + 0.712g oxygen = 1.483g total.
%Cr = (0.771/1.483)*100 = 51.99%
%O = (0.712/1.483)*100 = 48.01%
CrO, CrO2, Cr2O3, don't have these percentages.
If the equation is
2Cr + O2 ==> 2CrO (eqn 1)
Cr + O2 ==> CrO2 (eqn 2)
2Cr + 3O2 ==> 2CrO3 (eqn 3)
4Cr + 3O2 ==> 2Cr2O3 (eqn 4)
2Cr + 5O2 ==> 2CrO5
eqn 1. 0.771 g Cr uses 0.237g O2.
eqn 2. 0.771 g Cr uses 0.474 g O2.
eqn 3. 0.771 g Cr uses 0.712 g O2.
eqn 4. 0.771 g Cr uses 0.355 g O2.
eqn 5. 0.771 g Cr uses 1.186 g O2.
Equation 3 is the only one that fits.There may be a tricky part of Cr chemistry that I don't know but I'll stick with my first answer.
0.771g Cr + 0.712g oxygen = 1.483g total.
%Cr = (0.771/1.483)*100 = 51.99%
%O = (0.712/1.483)*100 = 48.01%
CrO, CrO2, Cr2O3, don't have these percentages.
If the equation is
2Cr + O2 ==> 2CrO (eqn 1)
Cr + O2 ==> CrO2 (eqn 2)
2Cr + 3O2 ==> 2CrO3 (eqn 3)
4Cr + 3O2 ==> 2Cr2O3 (eqn 4)
2Cr + 5O2 ==> 2CrO5
eqn 1. 0.771 g Cr uses 0.237g O2.
eqn 2. 0.771 g Cr uses 0.474 g O2.
eqn 3. 0.771 g Cr uses 0.712 g O2.
eqn 4. 0.771 g Cr uses 0.355 g O2.
eqn 5. 0.771 g Cr uses 1.186 g O2.
Equation 3 is the only one that fits.There may be a tricky part of Cr chemistry that I don't know but I'll stick with my first answer.