Is there any answers to choose from?
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Answer:
0.250 moles of MgO are produced when 0.250 mol of Mg reacts completely with O₂
Explanation:
In first place, the balanced reaction between Mg and O₂ is:
2 Mg + O₂ ⇒ 2 MgO
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactants and products participate in the reaction:
- Mg: 2 moles
- O₂: 1 mole
- MgO: 2 moles
Then you can apply the following rule of three: if by reaction stoichiometry 2 moles of Mg produce 2 moles of MgO, 0.250 moles of Mg, how many moles of MgO will they form?

moles of MgO= 0.250
<u><em>0.250 moles of MgO are produced when 0.250 mol of Mg reacts completely with O₂</em></u>
Correct me if I wrong but I think it's "c"
Answer:
see explanation below
Explanation:
First to all, this is a redox reaction, and the reaction taking place is the following:
2KMnO4 + 3H2SO4 + 5H2O2 -----> 2MnSO4 + K2SO4 + 8H2O + 5O2
According to this reaction, we can see that the mole ratio between the peroxide and the permangante is 5:2. Therefore, if the titration required 21.3 mL to reach the equivalence point, then, the moles would be:
MhVh = MpVp
h would be the hydrogen peroxide, and p the permanganate.
But like it was stated before, the mole ratio is 5:2 so:
5MhVh = 2MpVp
Replacing moles:
5nh = 2MpVp
Now, we just have to replace the given data:
nh = 2MpVp/5
nh = 2 * 1.68 * 0.0213 / 5
nh = 0.0143 moles
Now to get the mass, we just need the molecular mass of the peroxide:
MM = 2*1 + 2*16 = 34 g/mol
Finally the mass:
m = 0.0143 * 34
m = 0.4862 g