2NaClO₃ → 2NaCl + 3O₂
mole ratio of NaClO₃ to O₂ is 2 : 3
∴ if moles of NaClO₃ = 12 mol
then moles of O₂ =
= 18 mol
Mass of O₂ = mol of O₂ × molar mass of O₂
= 18 mol × 16 g/mol
= 288 g
So I wasn't sure which equation to use since you did not specify so I just used the decomposition reaction. If you should have used another reaction then just follow the same steps and you'll get your answer.
Answer:If we have [H+][OH-]= Kw = 1.0 x 10^-14
Then [H+]= Kw/ [OH-]= 1.0x 10^-14/ 1 x 10^-11 =1 x 10^-3 mol/L
And here is the solution - as you can see it is an acidic one :
pH = - log [H+]= - log 1 x 10^-3 = 3 < 7
Explanation:
Hi the answer is actually B
Answer:
85.34g of NH3
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Step 2:
Determination of the number of moles of NH3 produced by the reaction of 2.51 moles of N2. This is illustrated below:
From the balanced equation above,
1 mole of N2 reacted to produce 2 moles of NH3.
Therefore, 2.51 moles of N2 will react to produce = (2.51 x 2)/1 = 5.02 moles of NH3.
Therefore, 5.02 moles of NH3 is produced from the reaction.
Step 3:
Conversion of 5.02 moles of NH3 to grams. This is illustrated below:
Molar mass of NH3 = 14 + (3x1) = 17g/mol
Number of mole of NH3 = 5.02 moles
Mass of NH3 =..?
Mass = mole x molar Mass
Mass of NH3 = 5.02 x 17
Mass of NH3 = 85.34g
Therefore, 85.34g of NH3 is produced.
N= m/v n=0.077/.200 = 0.385