Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
So it's good to map out what you know you have and work from there:
We have two liter measurements and one mole measurement, and we need to find the moles.
For this problem, think of it this way: 46 liters of gas = 1.4 moles.
If one side changes, the other has to as well (if the liters decrease, the moles decrease. if the liters increase, so do the moles.) What you can do is put this into a fraction:
<span><u>1.4 moles</u></span>
46 L <span> </span>
if we know that each liter of gas is equal to x amount of moles, we know that 11.5 liters equals some amount of moles. You can put this into a fraction too, and make it equal to the other fraction:
<span><u>1.4 moles</u></span> = <u>x moles</u>
46 L 11.5 L
Then get your calculator out and do some algebra.
11.5 * (1.4/46) = x
The answer should come out to be: 0.35 moles
