The balanced reaction equation for the combustion of butane is as follows;
C₄H₁₀ + 13/2O₂ ---> 4CO₂ + 5H₂O
the limiting reactant in this reaction is C₄H₁₀ This means that all the butane moles are consumed and amount of product formed depends on the amount of C₄H₁₀ used up.
stoichiometry of C₄H₁₀ to H₂O is 1:5
mass of butane used - 6.97 g
number of moles - 6.97 g / 58 g/mol = 0.12 mol
then the number of water moles produced - 0.12 mol x 5 = 0.6 mol
Therefore mass of water produced - 0.6 mol x 18 g/mol = 10.8 g
1 mole ------------- 6.02 x 10²³ molecules
( moles ) ---------- 2.0 x 10²² molecules
moles = ( 2.0 x 10²² ) x 1 / 6.02 x 10²³
moles = 2.0 x 10²² / 6.02 x 10²³
= 0.03 moles of water
hope this helps!
Atoms are made up of Protons and neutrons in the nucleus, and then rings of electrons on the outer shells! so in short, protons, neutrons and electrons. :)
Answer:
0.39 mol
Explanation:
Considering the ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
At same volume, for two situations, the above equation can be written as:-
Given ,
n₁ = 1.50 mol
n₂ = ?
P₁ = 3.75 atm
P₂ = 0.998 atm
T₁ = 21.7 ºC
T₂ = 28.1 ºC
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (21.7 + 273.15) K = 294.85 K
T₂ = (28.1 + 273.15) K = 301.25 K
Using above equation as:
Solving for n₂ , we get:
n₂ = 0.39 mol