I think it is a and c but i’m not sure
Answer:
15.9 g
Explanation:
(Take the atomic mass of C=12.0, H=1.0, O=16.0)
no. of moles = mass / molar mass
no. of moles of octane used = 11.2 / (12.0x8 + 1x18)
= 0.0982456 mol
Since oxygen is in excess and octane is the limiting reagent, the no. of moles of H2O depends on the no. of moles of octane used.
From the balanced equation, the mole ratio of octane : water = 2:18 = 1: 9,
so this means, one mole of octane produced 9 moles of water.
Using this ratio, we can deduce that (y is the no. of moles of water produced):

y = 0.0982456x9
y= 0.88421 mol
Since mass = no. of moles x molar mass,
mass of water produced = 0.88421 x (1.0x2+16.0)
=15.9 g
The solution for the problem is:
First, use the concentration of the volume of the thing you
know to compute for the moles of that substance. Then, use the coefficient in
the balanced equation to relay moles of that to moles of anything else in the
chemical equation. Lastly, translate moles into mass by means of its molar
mass, or into a concentration using the volume.
Applying what I have said earlier:
0.0133 L X 1.68 mol/L = 0.0223 mol KMnO4 X (1 mol H2O2 / 2
mol KMnO4) = 0.0112 mol H2O2
Mass H2O2 = 0.0112 mol H2O2 X 34.0 g/mol = 0.380 grams H2O2
Answer:
30 g. 8'C
Explanation:
the 10 g is plus the fact that I have to help you get the chance to get the answer