First write out the balanced equation. 3Cu+2Ag(NO3)3=2Ag+3Cu(NO3)2
Then convert copper from grams to moles
15 g*1 mol cu/63.54 g= 15/63.54 mol cu
Then use the mole ratio to convert Moles Cu to Moles Ag
15/63.54 moles Cu* 2 moles Ag/3 moles Cu
The final awnser is (15*2)/(63.54*3) moles Ag =0.157 moles Ag. If the question wants the answer in grams, convert from moles Ag to grams Ag.
0.157 moles Ag*107.87 g Ag/ mol Ag=16.98 g Ag
<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:
None of these are correct, because there is no way to balance this equation, but I hope these steps help you figure out your answer.
Explanation:
Count out the single amounts of elements you have on both sides of the equation. To be balanced, you need to have the exact same for each element.
Before balanced Left side.
Cl-2
O-8
H-2
Before balanced right side.
H-1
Cl-1
O-3
That means we need to increase Hydrogen, Chlorine and Oxygen on the right for sure and see how that affects the equation. You can keep adding the Coefficients until the # of elements begin to match on each side.
(I tried to balance this equation, it doesn't work, there is too much on the reactants side for what the product is.)
Explanation:
Fe. O
72.40/ 56 27.60/16
____________________
1.29/1.29 1.725/1.29
_____________________
1. :. 1
<h3>Emperical formula = FeO</h3>
C
I have had this question on a test before!! Hope this helps
