Question

Consider the following balanced equation: 13O2(g) + 2C4H10(g) → 8CO2(g) + 10H2O(l)

If 2.90×102 moles of O2(g) and 27.1 moles of C4H10(g) are allowed to react, what is the theoretical yield of CO2(g)?

Answer #1

13 moles of O_{2} reacts with 2 moles of C4H10

1 mole of O2 reacts with (2/13) moles of C4H10

2.90 X 10^{2} moles of O2 reacts with (2 X 2.90 X
100/13) =44.61 moles of C4H10

But only 27.1 moles of C4H10 are available.So, C4H10 is the limiting reagent.So,we will calculate the yield with the given moles of C4H10.

2 moles of C4H10 gives 8 moles of CO2

1 mole of C4H10 gives (8/2) moles of CO2

27.1 moles of C4H10 gives (8 X 27.1/2) =108.4 moles of CO2

Now,

No of moles=given weight/molecular weight

108.4 mol= given weight/44 g/mol

Given weight=108.4 X 44= 4769.6 g

write the balanced chemical equation for when butane
burns
I think it is 2c4h10+13o2 = 8co2 + 10h2o now how do I find the
change in enthalpy

2C4H10(g)+13O2(g)→10H2O(g)+8CO2(g) Calculate the mass of water
produced when 4.16 g of butane reacts with excess oxygen.

2C4H10(g)+13 O2(g)--> 10H2O(g)+8CO2(g)
A. calculate the mass of water produced when 6.30g of
butane (C4H10) reacts with excess oxygen.
B. calculate the mass of butane needed to produce
47.2g of carbon dioxide

Mass C4H10
Mass O2
Mass CO2
Mass H2O
1.31 g
5.72g
11.12g
8.84g
222 mg
148mg
Consider the following balanced equation for the combustion of
butane, a fuel often used in lighters.
2C4H10(g)+13O2(g)?8CO2(g)+10H2O(g)
Complete the following table, showing the appropriate masses of
reactants and products. If the mass of a reactant is provided, fill
in the mass of other reactants required to completely react with
the given mass, as well as the mass of each product formed. If the
mass...

Butane, C4H10 , is a component of natural gas that is used as
fuel for cigarette lighters. The balanced equation of the complete
combustion of butane is
2C4H10(g)+13O2(g)→8CO2(g)+10H2O(l)
At 1.00 atm and 23 ∘C , what is the volume of carbon dioxide
formed by the combustion of 1.60 g of butane?
Express your answer with the appropriate units.
volume of CO2 =

Butane, C4H10, is a component of natural gas that is used as
fuel for cigarette lighters. The balanced equation of the complete
combustion of butane is 2C4H10(g)+13O2(g)→8CO2(g)+10H2O(l) At 1.01
bar and 23 ∘C, what is the volume of carbon dioxide formed by the
combustion of 2.60 g of butane?

Butane, C4H10, is a component of natural gas that is used as
fuel for cigarette lighters. The balanced equation of the complete
combustion of butane is 2C4H10(g)+13O2(g)→8CO2(g)+10H2O(l) At 1.00
atm and 23 ∘C, what is the volume of carbon dioxide formed by the
combustion of 1.60 g of butane?

Butane undergoes combustion when it reacts with oxygen to
produce carbon dioxide and water.
2C4H10(g)+13O2(g)⟶Δ8CO2(g)+10H2O(g)
Part A
What volume, in liters, of oxygen is needed to react with 54.9 g
of butane at 0.84 atm atm and 29 ∘C?
Express your answer with the appropriate units.

Thermochemical Equations 07a (kJ)
Consider the following thermochemical equation for the
combustion of butane.
2C4H10(g)+15O2(g)→8CO2(g)+10H2O(g)ΔH∘rxn=−5314.6kJ
Part A
Calculate the heat associated with the consumption of 1.158 mol
of O2 in this reaction.
Use the correct sign for q
q =
+
kJ
SubmitMy AnswersGive
Up
Incorrect; Try Again; 6 attempts remaining
Part B
Calculate the heat associated with combustion of 29.46 g of
butane.
Use the correct sign for q
q =
kJ
SubmitMy AnswersGive
Up
Part C
Calculate the...

The ideal gas law
PV=nRT
relates pressure P, volume V, temperature
T, and number of moles of a gas, n. The gas
constant Requals 0.08206 L⋅atm/(K⋅mol) or 8.3145
J/(K⋅mol). The equation can be rearranged as follows to solve for
n:
n=PVRT
This equation is useful when dealing with gaseous reactions
because stoichiometric calculations involve mole ratios.
A)When heated, calcium carbonate decomposes to yield calcium
oxide and carbon dioxide gas via the reaction
CaCO3(s)→CaO(s)+CO2(g)
What is the mass of calcium carbonate...

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