Answer:
2C3H8O + 9O2 ==> 6CO2 + 8H2O ... balanced equation
moles propanol = 5.26 g x 1 mol/60.1 g = 0.0875 moles
moles O2 = 31.8 g x 1 mol/31.9 g = 0.997 moles O2
Propanol is limiting based on the mol ratio in balance equation of 2 : 9
To find mass of O2 (excess reagent) left over, we will first find moles O2 used up.
moles O2 used = 0.0875 mol propanol x 9 mol O2/2 mol propanol = 0.394 moles O2 used
moles O2 left over = 0.997 mol - 0.394 mol = 0.603 mol O2 left
mass O2 left = 0.603 mol O2 x 32 g/mol = 19.3 g O2 left over
Hi,
Homogeneous mixture 2
Heterogeneous 3
Mixture 1
Solution 5
Compound 4
Within the core of the Sun, temperatures and pressures are high enough to fuse hydrogen atoms into helium, which is the Sun's main form of energy production. Assuming there was a slight mistake in where you have copied the results here the correct answer is the third option.
Hope this helps!
Answer:
11.3 g.
Explanation:
Hello there!
In this case, since the combustion of butane is:
Thus, since there is a 1:5 mole ratio between butane and water, we obtain the following mass of water:
Therefore, the resulting mass of water is:
Best regards!
