5. When propane is reacted in the presence of oxygen gas, the products of this combustion reaction are: a. CH2 + H2O b. C + H2 c. CO2 + H2O d. CO2 + H2
Answers
Answer:-
Propane gas (C3H8) burns with oxygen gas and products carbon dioxide gas, liquid water and
energy.
a) Write the balanced equation.
b) How many moles are in 66,0 g of propane?
c) How many moles of oxygen would be needed?
d) Calculate the mass of CO2 would be produced when 66,0 g of propane are burned.
a) First, we write the chemical reaction and we calculate the relative molecular mass (or
atomic mass if it's an element) of each substance in the reaction:
C3H8 + O2 → CO2 + H2O
C3H8 Ma = 3 · 12,0 + 8 · 1,0 = 44,0 36,5 u (1
)
O2 Mm= 2 · 16,0 = 32,0 u
CO2 Mm= 1 · 12,0 + 2 · 16,0 = 44,0 u
H2O Mm= 2 · 1,0 + 1 · 16,0 = 18,0 u
Mm 44,0 32,0 44,0 18,0
C3H8 + O2 → CO2 H2O
b) Second, we calculate the amount of substance (n) with the given data in the statement.
n = 66,0 g C3H8 · 1 mol C3H8 / 44,0 g C3H8 = 1,5 moles C3H8
c) Third, we balance the chemical equation.
Chemical reaction: C3H8 + O2 → CO2 + H2O
C 3 1
H 8 2
O 2 2 1
Carbon, hydrogen and oxygen aren't balanced. We balance carbon and hydrogen first:
C3H8 + O2 → 3 CO2 + 4 H2O
C 3 3
H 8 8
O 2 6 4
Oxygen is not balanced:
C3H8 + 5 O2 → 3 CO2 + 4 H2O
C 3 3
H 8 8
O 10 6 4
Balanced equation: C3H8 + 5 O2 → 3 CO2 + 4 H2O
1 unified atomic mass unit, symbol (u)
d) Then, we collect all the information:
Mm 44,0 32,0 44,0 18,0
Stoichiometry 1 mol 5 moles 3 moles 4 moles
C3H8 + 5 O2 → 3 CO2 4 H2O
Amount of
substance
1,5 moles x moles y moles z moles
e) Fourth, we calculate the amount of the substances (n) in the chemical reaction:
x moles O2 = 1,5 moles C3H8 · 5 moles O2 / 1 mol C3H8 = 7,5 moles O2
y moles CO2 = 1,5 moles C3H8 · 3 moles CO2 / 1 mol C3H8 = 4,5 moles CO2
z moles H2O = 1,5 moles C3H8 · 4 mol H2O / 1 mol C3H8 = 6,0 moles H2O
f) We collect the information again:
Mm 44,0 32,0 44,0 18,0
Stoichiometry 1 mol 5 moles 3 moles 4 moles
C3H8 + 5 O2 → 3 CO2 4 H2O
Amount of
substance
1,5 moles 7,5 moles 4,5 moles 6,0 moles
g) Then, we can answer the questions in the statement by using the amounts of substances:
a) Write the balanced equation.
(step c) Balanced equation: C3H8 + 5 O2 → 3 CO2 + 4 H2O
b) How many moles are in 66,0 g of propane?
(step b) n = 66,0 g C3H8 · 1 mol C3H8 / 44,0 g C3H8 = 1,5 moles C3H8
c) How many moles of oxygen would be needed?
(step e) x moles O2 = 1,5 moles C3H8 · 5 moles O2 / 1 mol C3H8 = 7,5 moles O2
d) Calculate the mass of CO2 would be produced when 66,0 g of propane are burned.
(step e) There will be produced 4,5 moles of CO2.
The mass is: 4,5 moles CO2 · 44,0 g CO2 /1 mol CO2 = 198,0 g CO2