Answer:
80.27%
Explanation:
Let's consider the following balanced equation.
2 Fe³⁺(aq) + Sn²⁺(aq) ⇒ 2Fe²⁺(aq) + Sn⁴⁺(aq)
First, we have to calculate the moles of Sn²⁺ that react.

We also know the following relations:
- According to the balanced equation, 1 mole of Sn²⁺ reacts with 2 moles of Fe³⁺.
- 1 mole of Fe³⁺ is oxidized from 1 mole of Fe.
- The molar mass of Fe is 55.84 g/mol.
Then, for 1.348 × 10⁻3 moles of Sn²⁺:

If there are 0.1505 g of Fe in a 0.1875 g sample, the mass percentage of Fe is:

The formula for water is H2O so there would have to be two Hyrdogens and one oxygen. Therefore it would be 4g of Hydrogen and 16g of Oxygen leaving you with 20g.
The answer is D.
Hope this helps :) ~
Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :
[butane]=1.0 M , [isobutane]=2.5 M
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?
Answer:
The equilibrium concentration of each gas:
[Butane] = 1.14 M
[isobutane] = 2.86 M
Explanation:
Butane ⇄ Isobutane
At equilibrium
1.0 M 2.5 M
After addition of 0.50 M of butane:
(1.0 + 0.50) M -
After equilibrium reestablishes:
(1.50-x)M (2.5+x)
The equilibrium expression will wriiten as:
![K_c=\frac{[Isobutane]}{[Butane]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BIsobutane%5D%7D%7B%5BButane%5D%7D)

x = 0.36 M
The equilibrium concentration of each gas:
[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M
[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M