2KClO3 --> 2KClO2 + O2
12 6 (moles)
The ratio of KClO3 and O2 is 2:1. This means 2 moles of KClO3 can create 1 mole of O2. So 12 moles of KClO3 will create 6 moles of O2.
Answer : The concentration of ![[Fe(SCN)]^{2+}](https://tex.z-dn.net/?f=%5BFe%28SCN%29%5D%5E%7B2%2B%7D)
is, 
Explanation :
When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is 
and 
is excess reagent.
First we have to calculate the moles of KSCN.
Moles of KSCN = Moles of 
= Moles of = 
Now we have to calculate the concentration of
![\text{Concentration of }[Fe(SCN)]^{2+}=\frac{\text{Moles of }[Fe(SCN)]^{2+}}{\text{Volume of solution}}](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%7D%7B%5Ctext%7BVolume%20of%20solution%7D%7D)
Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L
![\text{Concentration of }[Fe(SCN)]^{2+}=\frac{1.08\times 10^{-5}mol}{0.025L}=4.32\times 10^{-4}M](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%3D%5Cfrac%7B1.08%5Ctimes%2010%5E%7B-5%7Dmol%7D%7B0.025L%7D%3D4.32%5Ctimes%2010%5E%7B-4%7DM)
Thus, the concentration of is,
N(CH₃OH)=3,62·10²⁴/6·10²³ 1/mol = 6,033 mol
m(CH₃OH) = 6,033 mol · 32 g/mol (molar mass) = 193,06 g.
corrected question: A chemist adds 135mL of a 0.21M zinc nitrate solution to a reaction flask. Calculate the mass in grams of zinc nitrate the chemist has added to the flask. Round your answer to significant digits.
Answer:
5.37g
Explanation:
0.21M means ; 0.21mol/dm³
1dm³=1L , so we can say 0.21mol/L
if 0.21mol of Zinc nitrate is contained in 1L of water
x will be contained in 135mL of water
x= 0.21*135*10³/1
=0.02835moles
number of moles= mass/ molar mass
mass= number of moles *molar mas
molar mass of Zn(NO₃)₂=189.36 g/mol
mass= 0.02835 *189.36
mass=5.37g
what is the pOH of the solution? 10.25 is the answer
