Question

14.7 g of butane (58.12 g/mol) undergoes combustion according to the following equation. What pressure of carbon dioxide in atm is produced at 318 K in a 1.58 L flask. 2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (g)

Answer #1

Molar mass of C4H10 = 58.12 g/mol

mass of C4H10 = 14.7 g

mol of C4H10 = (mass)/(molar mass)

= 14.7/58.12

= 0.2529 mol

From balanced chemical reaction, we see that

when 2 mol of C4H10 reacts, 8 mol of CO2 is formed

mol of CO2 formed = (8/2)* moles of C4H10

= (8/2)*0.2529

= 1.0117 mol

we have:

V = 1.58 L

n = 1.0117 mol

T = 318.0 K

we have below equation to be used:

P * V = n*R*T

P * 1.58 L = 1.0117 mol* 0.08206 atm.L/mol.K * 318 K

P = 16.7 atm

Answer: 16.7 atm

19.4 g of butane (58.12 g/mol) undergoes combustion according to
the following equation. What pressure of carbon dioxide in atm is
produced at 309 K in a 1.15 L flask.
2 C4H10(g) + 13 O2 (g) → 8
CO2 (g) + 10 H2O (g)

(a) Consider the combustion of butane, given below: 2 C4H10(g) +
13 O2(g) 8 CO2(g) + 10 H2O(g) If C4H10(g) is decreasing at the rate
of 0.850 mol/s, what are the rates of change of O2(g), CO2(g), and
H2O(g)?
O2(g)/t = mol/s
CO2(g)/t = mol/s
H2O(g)/t = mol/s
(b) The decomposition reaction given below: 2 IF5(g) 1 I2(g) + 5
F2(g) is carried out in a closed reaction vessel. If the partial
pressure of IF5(g) is decreasing at the rate...

Butane, C4H10, is a component of natural gas that is used as
fuel for cigarette lighters. The balanced equation of the complete
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atm and 23 ∘C, what is the volume of carbon dioxide formed by the
combustion of 2.80 g of butane? what is the volume of CO2?The ideal
gas law
PV=nRT
relates pressure P, volume V, temperature
T, and number of moles of a gas, n. The gas
constant R equals 0.08206...

Butane, the fuel used in cigarette lighters, burns according to
the equation: 2 C4H10 (g) + 13 O2 (g) 8 CO2 (g) + 10 H2O(g) H =
– 5316 kJ a) Calculate the mass of oxygen that must react in order
for this reaction to generate 2150 kJ of heat b) Calculate the
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18. Butane gas burns according to the following exothermic
reaction:
C4H10 (g) + 13/2 O2 (g) → 4 CO2 (g) + 5 H2O (g) ∆H°rxn = -
2877.1 kJ
a) If 25.0 g of butane were burned, how much energy would be
released?
b) If the reaction of 25.0 g of butane produced a volume change
of 15.4 L against an external pressure of 748 mmHg, calculate the
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c) Calculate the change in internal energy (∆E)...

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...

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 =

A) Butane combusts according to the equation: C4H10(g)+132O2(g)→4CO2(g)+5H2O(g)
ΔHrxn=−2658kJ
- What mass of butane in grams is
necessary to produce 1.1×103 kJ kJ of heat? Express your answer to two significant figures
and include the appropriate units.
- What mass of CO2 is
produced? Express your answer
to two significant figures and include the appropriate
units.
B) Charcoal is primarily carbon. What mass of CO2
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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.
Part A
When heated, calcium carbonate decomposes to yield calcium oxide
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2C8H18(l) + 25O2(g) 16CO2(g) +
18H2O(l) ∆H°rxn = −10,800
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Given that ∆H°f[CO2(g)] = −394 kJ/mol and ∆H°f[H2O(l)] = −286
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octane.
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b.326 kJ/mol
c.210 kJ/mol
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