Moles of Carbon dioxide : 13.5
<h3>Further explanation</h3>
Given
4.5 moles propane
Required
moles of Carbon dioxide
Solution
Reaction(Combustion of Propane) :
<em>C₃H₈(g)+5O₂(g)→3CO₂(g)+4H₂O(g)</em>
In a chemical equation, the reaction coefficient shows the mole ratio of the reacting compounds (either in reactants or products)
From the equation, mol ratio of Propane(C₃H₈) : Carbon dioxide(CO₂) : 1 : 3, so mol CO₂ :
= 3/1 x mol Propane
= 3 x 4.5 moles
= 13.5 moles
Answer:
#P: 15
#N: 16
Mass Number: 30.973762u (You can round it if you want)
Isotope symbol: P
Explanation:
I hope this helps :)
From the equation;
P1V1=P2V2
V2=P1V1÷P2
V2=29.0×13.0÷3.50
V2=107.71L
therefore V2=107.7L
In the given question,
HgxOy (2.50 g) = xHg (2.405 g) + yO
Here, mass of oxygen = 2.50 - 2.405
= 0.095 g
The moles of O₂ = 0.095 / 16 = 0.00594 moles
Moles of Hg = 2.405 / 200.60 = 0.01198 moles
Thus, the ratio of Hg : O is,
0.00594 : 0.01198 or 1 : 2
As the ratio is 1 : 2, thus, the empirical formula is Hg₂O