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
The answer is 1.5e+6. Hope this helped!
Hydrogen sulfide = hidrogen + sulfur
6.500 g
a) 0.384 g + x
=> 6.500 = 0.384 + x => x = 6.500 - 0.384 = 6.116 g
Answer: 6.116 g of sulfur must be obtained
b) this experiment demonstrate the conservation of mass.
c) Dalton's atomic model states that the atoms cannot be created, split or be destroyed, and so in a chemical reaction the atoms rearrange but the number of each type of atoms remain constant, so the mass of each type of atoms and the total mass remain constant.
Standard temperature is 273 K
Standard pressure is 1 atm
We use the ideal gas equation to find out density of nitrogen gas in g/L
Ideal gas equation:
Molar mass of 
Pressure = 1 atm
Temperature = 273 K
= 1.25 g/L
Therefore, density of nitrogen gas at STP is 1.25 g/L
1.1 moles of C10H8 is the limiting reagent in the reaction between reaction C10H8 and O2
.
C10H8 + 12O2 ----> 10CO2 + 4H2O
C10H8 is the limiting reagent since 1.1 moles of C10H8 is totally consumed during the reaction
