Question

The ideal gas law

*P**V*=*n**R**T*

relates pressure *P*, volume *V*, temperature
*T*, and number of moles of a gas, *n*. The gas
constant *R*equals 0.08206 L⋅atm/(K⋅mol) or 8.3145
J/(K⋅mol). The equation can be rearranged as follows to solve for
*n*:

*n*=*P**V**R**T*

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 needed to produce 75.0 L of carbon dioxide at STP?

B)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 1.20 g of butane?

Answer #1

A) first calculate mole of carbon dioxide by using ideal gas equation

Ideal gas equation

PV = nRT where, P = atm prrssure = 1 atm,

V = volume in Liter = 70 L

n = number of mole = ?

R = 0.08205L atm mol^{-1} K^{-1}
=Proportionality constant = gas constant,

T = Temperature in K = 25^{0}C = 273.15+ 25 = 298.15
K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (170) / (0.08205 298.15) = 2.86 mole

according to reaction to produce 1 mole of carbon dioxide require 1 mole calcium carbonate then to produce 2.86 mole of carbon dioxide require 2.86 mole of calcium carbonate

molar mass of calcium carbonate = 100.0869 gm/mol then 2.86 mole = 100.0869 2.86 = 286.25 gm

286.25 gm calcium carbonate require to produce 70 L carbon dioxide at STP

B) molar mass of butane = 58.12 gm/mol then 2 mole of butane = 116.24 gm

molar mass of carbon dioxide = 44.01 gm/mol then 8 mole carbon dioxide = 352.08 gm

According to reaction 2 mole of butane produce 8 mole of carbon dioxide that mean 116.24 gm butane produce 352.08 gm of carbon dioxide then 0.20 gm butane produce

352.08 1.20 / 116.24 = 3.63 gm of carbon dioxide

molar mass of carbon dioxide = 44.01 gm/mol then 3.63 gm = 3.63/ 44.01 = 0.0825 mole of carbon dioxide

caculate volume of carbon dioxide by using ideal gas equation

We know that PV = nRT

V = nRT/P

n = 0.0825 mole,

T = 23^{0}C = 23+273.15 = 296.15K,

P= 1 atm,

R = 0.08205 L atm mol^{-1} K^{-1} ( R = gas
constant)

V = ?

Substitute these value in above equation.

V = (0.0825 0.08205 296.15)/ 1 = 2.0046 L

At 1.00 atm and 23 ∘C, 2.0046 L volume of carbon dioxide formed by the combustion of 1.20 g of butane

± Stoichiometric Relationships with Gases
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.
Part A
When heated, calcium carbonate decomposes to yield calcium oxide
and carbon dioxide gas via the reaction
CaCO3(s)→CaO(s)+CO2(g)...

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 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, 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, 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?

1.) Imagine that you have a 7.00 L gas tank and a 4.00 L gas
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acetylene to use in conjunction with your welding torch. If you
fill the larger tank with oxygen to a pressure of 155 atm , to what
pressure should you fill the acetylene tank to ensure that you run
out of each gas at the same time? Assume ideal behavior for all
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1. The Ideal Gas Law equation is PV=nRT. Define each variable
and include the unit necessary for each variable in the
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2. What is the pressure in a vessel of 4.5g of O2 gas
that takes up a volume of 4.2 L and is at a temperature of 38.2
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3. A vessel of N2 gas was originally at a pressure of
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14.7 g of butane (58.12 g/mol) undergoes combustion according to
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CO2 (g) + 10 H2O (g)

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)

Consider the Ideal Gas Law, which states that PV = nRT, where P
is the pressure, V is the volume, T is the temperature, and n is
the number of moles of a gas sample, and R is a constant. (a)
Assume a sample of 1 mole of a gas is in a expandable container
where temperature and pressure are allowed to vary. Solve this
equation for V = f(P,T).
(b) Determine ∂V/dP and interpret the result. In particular,
describe...

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