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.
· 7 years ago