<span>1.40 x 10^5 kilograms of calcium oxide
The reaction looks like
SO2 + CaO => CaSO3
First, determine the mass of sulfur in the coal
5.00 x 10^6 * 1.60 x 10^-2 = 8.00 x 10^4
Now lookup the atomic weights of Sulfur, Calcium, and Oxygen.
Sulfur = 32.065
Calcium = 40.078
Oxygen = 15.999
Calculate the molar mass of CaO
CaO = 40.078 + 15.999 = 56.077
Since 1 atom of sulfur makes 1 atom of sulfur dioxide, we don't need the molar mass of sulfur dioxide. We merely need the number of moles of sulfur we're burning. divide the mass of sulfur by the atomic weight.
8.00 x 10^4 / 32.065 = 2.49 x 10^3 moles
Since 1 molecule of sulfur dioxide is reacted with 1 molecule of calcium oxide, just multiply the number of moles needed by the molar mass
2.49 x 10^3 * 56.077 = 1.40 x 10^5
So you need to use 1.40 x 10^5 kilograms of calcium oxide per day to treat the sulfur dioxide generated by burning 5.00 x 10^6 kilograms of coal with 1.60% sulfur.</span>
Answer:
7 hours
Explanation: step 1. 160x4=640
step 2. 1180-640=540
step 3. 540÷180=3
step 4. 3+4=7
Explanation:
Percentage composition = 2/18 = 11.11%.
Answer:
Amount of excess Carbon (ii) oxide left over = 23.75 g
Explanation:
Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂
Molar mass of Fe₂O₃ = 160 g/mol;
Molar mass of Carbon (ii) oxide = 28 g/mol
From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g) of carbon (ii) oxide
450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide
Therefore the excess reactant is carbon (ii) oxide.
Amount of excess Carbon (ii) oxide left over = 260 - 236.25
Amount of excess Carbon (ii) oxide left over = 23.75 g
