Esay 1,050 pts! 1) Acetylene torches are used for welding. These torches use a m

Question

Esay 1,050 pts!

1) Acetylene torches are used for welding. These torches use a mixture of acetylene gas, C2H2, and oxygen gas, O2 to produce the following combustion reaction:

2C2H2(g)+5O2(g)?4CO2(g)+2H2O(g

Imagine that you have a 5.00L gas tank and a 2.00L gas tank. You need to fill one tank with oxygen and the other with acetylene to use in conjunction with your welding torch. If you fill the larger tank with oxygen to a pressure of 125atm , 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 gases.

(781 atm is wrong and 1562 atm is wrong too)

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2) (use pv=nRT)

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 73.0L of carbon dioxide at STP?

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3) (use pv=nRT)

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 3.20g of butane?

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4)

Consider the chemical reaction:
2H2O(l)?2H2(g)+O2(g)

What mass of H2O is required to form 1.3L of O2 at a temperature of 325K and a pressure of 0.966atm ?

Explanation / Answer

1)

First thing to understand is how you can manupulate the Ideal Gas Law to your needs. Since PV = nRT, at a constant temperature, PV/n = RT = constant. That means

P1V1/n1 = P2V2/n2!

If we say that subscript 1 = oxygen and subscript 2 = acetylene, then for acetylene, P2 = (P1V1n2) / (V2n1).

Next, we realize from the reaction equation given above that 2 moles of acetylene (n2) react with 5 moles of oxygen (n1). We can plug that right into our equation for P2, along with the rest of the conditions given in the question:

P2 = (P1V1n2) / (V2n1) = [(125 atm)*(5.00L)*(2 moles)] / [(2.00 L)*(5 moles)] = 125 atm

(2)

_____Numbers ALMOST always have unites! Presumably 73.0 has liters as its units

_____1 mole CaCO3 ---> 1 mole CO2

moles CO2 = 73.0 liters / 22.4 liters/mole at STP = 3.26 moles

g CaCO3 = (MW CaCO3 g/mole CaCO3) * (1 moles CaCO3/mole CO2) * x moles CO2 = ??

(3)

Use the ideal gas law.

pV = nRT

V = nRT/p

the number of moles of butane combusted is (3.20 g / 58.1222 g/mol) = 0.055 moles.
The number of moles of CO2 is four times the number of moles of butane, so n = 0.22 moles.

T = 23 + 273.15 = 296.15 K
R = 0.08206 (L atm) / (mol K)
p = 1 atm

so V = 0.22*0.08206*296.16/1 = 5.347 L

About 5.4 L of CO2 gas are formed.

(4)

Use the IDEAL GAS Eq'n

PV = nRT

P = 0.966 atm = 97879.95 Pa
V = 1.3L = 1.3 x 10^-3 m^3
R = 8.314 J k^-1 mol^-1
T = 325K

n = PV/RT
n = 97879.95 x 1.3 x 10^-3 / 8.314 x 325
n = 0.047 mol(O2) (Equivalent to '1' in the reaction eq'n)

n = 0.047 x 2 = 0.094 mol(H2O) (Equivalent to '2' in the reaction eq'n)

Is the answer required.

Mass(H2O) = 0.094 x 18 = 1.692 g is the mass required.

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