<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:
22.4L of one mole of any gas
or you can use PV=nRT
3.45*22.4=77.28
Explanation:
Answer:
The mass of
4.6
×
10
24
atoms of silver is approximately 820 g.
Explanation:
In order to determine the mass of a given number of atoms of an element, identify the equalities between moles of the element and atoms of the element, and between moles of the element and its molar mass.
1
mole atoms Ag=6.022xx10
23
atoms Ag
Molar mass of Ag =#"107.87 g/mol"#
Multiply the given atoms of silver by
1
mol Ag
6.022
×
23
atoms Ag
. Then multiply times the molar mass of silver.
4.6
×
10
24
atoms Ag
×
1
mol Ag
6.022
×
10
23
atoms Ag
×
107.87
g Ag
1
mol Ag
=
820 g Ag
Depends on the element it can by up to 3, 8, or maybe 16.
