Given the following unbalanced chemical reaction for the rusting or iron:
Fe(solid) + O2 (gas) + H2O (liquid) -> Fe2O3(H2O)2
If 1.5 liters of oxygen gas were add to excess water and iron at a temperature of 67 degrees C and a pressure of 1.02 atms, how much iron in grams actually rusts? How much product is made in grams?
Fe(solid) + O2 (gas) + H2O (liquid) -> Fe2O3(H2O)2
If 1.5 liters of oxygen gas were add to excess water and iron at a temperature of 67 degrees C and a pressure of 1.02 atms, how much iron in grams actually rusts? How much product is made in grams?
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First, you need to balance the equation. The balanced equation is:
4Fe + 3O2 + 4H2O -> 2Fe2O3(H2O)2
You are given the volume of the gas, the temperature and the pressure used. You can use good old Ideal Gas Equation.
PV=nRT
P = pressure (in atm)
V = volume (in L)
n = moles of gas
R = 0.08206 L.atm/mol.K
T = temperature (in K)
You plug in the values:
(1.02 atm)(1.5L) = n(0.08206 L.atm/mol.K)(67 + 273 K)
Formula transformation will get you the value of n (moles of gas, which is oxygen in this case)
n = 0.0548 moles O2
After, you use stoichiometry to determine the amount of Iron that rusts.
g Fe that rusts = 0.0548 mol O2 x (4 mol Fe / 3 mol O2) x (55.845 g Fe / mol Fe)
Cancelling the common units will leave you g Fe.
g Fe that rusts = 4.08 g
You also use stoichiometry toget the amount of rust (product) produced:
g rust (Fe2O3(H2O)2) = 0.0548 mol O2 x (2 mol Fe2O3(H2O)2 / 3 mol O2) x (195.73 g rust/mol rust)
g rust = 7.15 g of rust produced
4Fe + 3O2 + 4H2O -> 2Fe2O3(H2O)2
You are given the volume of the gas, the temperature and the pressure used. You can use good old Ideal Gas Equation.
PV=nRT
P = pressure (in atm)
V = volume (in L)
n = moles of gas
R = 0.08206 L.atm/mol.K
T = temperature (in K)
You plug in the values:
(1.02 atm)(1.5L) = n(0.08206 L.atm/mol.K)(67 + 273 K)
Formula transformation will get you the value of n (moles of gas, which is oxygen in this case)
n = 0.0548 moles O2
After, you use stoichiometry to determine the amount of Iron that rusts.
g Fe that rusts = 0.0548 mol O2 x (4 mol Fe / 3 mol O2) x (55.845 g Fe / mol Fe)
Cancelling the common units will leave you g Fe.
g Fe that rusts = 4.08 g
You also use stoichiometry toget the amount of rust (product) produced:
g rust (Fe2O3(H2O)2) = 0.0548 mol O2 x (2 mol Fe2O3(H2O)2 / 3 mol O2) x (195.73 g rust/mol rust)
g rust = 7.15 g of rust produced
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Basically, what this question is asking is for you to use the Ideal Gas Law to solve for the number of moles of oxygen gas and then use stoichiometry to solve for the rusted iron.
First, let's setup the Ideal Gas Law which is PV = nRT. Since we want moles, we can rearrange it to get n = PV/RT.
Now, plug in what we know:
P = 1.02atm
V = 1.5L
R = 0.08206L*atm/mol*K
T = 67C + 273.15 = 340.15K
n = (1.02atm)(1.5L)/(0.08206L*atm/mol*K)(340…
Now solve... After working the equation, we get 0.0548mol of oxygen gas.
Now to solve for grams of rust:
First off, balance the equation, the coefficients are [4, 3, 4, 2].
Since the ratio of oxygen to rust is 3:2, we can solve by multiplying the moles of oxygen * 2/3.
By solving this, we get 0.0365mol of rust, and converting to grams, we get 7.144g of rust formed.
First, let's setup the Ideal Gas Law which is PV = nRT. Since we want moles, we can rearrange it to get n = PV/RT.
Now, plug in what we know:
P = 1.02atm
V = 1.5L
R = 0.08206L*atm/mol*K
T = 67C + 273.15 = 340.15K
n = (1.02atm)(1.5L)/(0.08206L*atm/mol*K)(340…
Now solve... After working the equation, we get 0.0548mol of oxygen gas.
Now to solve for grams of rust:
First off, balance the equation, the coefficients are [4, 3, 4, 2].
Since the ratio of oxygen to rust is 3:2, we can solve by multiplying the moles of oxygen * 2/3.
By solving this, we get 0.0365mol of rust, and converting to grams, we get 7.144g of rust formed.