The given question is incomplete. The complete question is :
Gaseous butane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . If 1.31g of water is produced from the reaction of 4.65g of butane and 10.8g of oxygen gas, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.
Answer: 28.0 %
Explanation:
To calculate the moles :
According to stoichiometry :
13 moles of 
require 2 moles of butane
Thus 0.34 moles of will require=
of butane
Thus is the limiting reagent as it limits the formation of product and butane is the excess reagent.
As 13 moles of give = 10 moles of 
Thus 0.34 moles of give =
of
Mass of 
The percent yield of water is 28.0 %
Answer:
B. salt dissolved in water
Explanation:
a mixture is multiple substances mixed together
Answer:
Generally speaking, as the human population grows, our consumption of natural resources increases. More humans consume more freshwater, more land, more clothing, etc. Scientific and technological innovations mean that we are improving our efficiency at using and harvesting natural resources.
___________________________________
<h3>Please mark me as Brainlist</h3>
Answer:
The maximum amount of work that can be done by this system is -2.71 kJ/mol
Explanation:
Maximum amount of work denoted change in gibbs free energy 
during the reaction.
Equilibrium concentration of B = 0.357 M
So equilibrium concentration of A = (1-0.357) M = 0.643 M
So equilibrium constant at 253 K, ![K_{eq}= \frac{[B]}{[A]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%20%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D)
[A] and [B] represent equilibrium concentrations
When concentration of A = 0.867 M then B = (1-0.867) M = 0.133 M
So reaction quotient at this situation, 
We know, 
where R is gas constant and T is temperature in kelvin
Here R is 8.314 J/(mol.K), T is 253 K, Q is 0.153 and 
is 0.555
So, 
= -2710 J/mol
= -2.71 kJ/mol
