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
An not sure,, but maybe 24. don't take my word for it
This is what I got. Hope it helps :)
Answer:
0.07 g/s.
Explanation:
From the question given above, the following data were obtained:
Mass lost = 9.85 g
Time taken = 2 min 30 s
Mean rate =?
Next, we shall convert 2 min 30 s to seconds (s). This can be obtained as follow:
1 min = 60 s
Thus,
2 min = 2 × 60 = 120 s
Therefore,
2 min 30 s = 120 s + 30 s = 150 s
Finally, we shall determine the mean rate of the reaction. This can be obtained as illustrated below:
Mass lost = 9.85 g
Time taken = 150 s
Mean rate =?
Mean rate = mass lost / time taken
Mean rate = 9.85 / 150
Mean rate = 0.07 g/s
Therefore, the mean rate of the reaction is 0.07 g/s
On complete conversion (100% yield) 9.75 g of ethyl butyrate will be produced. Below is the solution.....
