1) Excess reagent
1 mol N2 / 3 mol H2
6.0 mol N2 *3 mol H2 / 1 mol N2 = 18 mol H2
18mol H2 > 12 mol H2 => H2 is limiting (you need 18 mol H2 to use all the 6 mol N2), then N2 is in excees.
12.0 mol H2 * 1mol N2/ 3 mol H2 = 4 mol N2 is the quantity that will react, then the excess is 6 mol N2 - 4 mol N2 = 2 mol N2
2) NH3 produced
12 mol H2 * [2 mol NH3 / 3 mol H2] = 8 mol NH3
Aslso, 4 mol N2 *[2molNH3 / 1 molN2] = 8 mol NH3, the same result.
3) Yield
80% * 8 mol NH3 = 6.4 mol NH3
Answer:
21.344%
Explanation:
For the given chemical reaction, 8 moles of the reactant should produce 4 moles of 
. However, 195 g of was produced instead. The molar mass of is 61.9789 g/mol.
Thus, the moles of produced = 195/61.9789 = 3.1462 moles
The percent error = [(Actual -Experiment)/Actual]*100%
The percent error = [(4.00 - 3.1462)/4.00]*100% = (0.85376/4.00)*100% = 21.344%
Answer:
1984
Explanation:
Given the formula;
0.693/t1/2 = 2.303/t log (Ao/A)
Where;
t1/2 = half life of the radioactive isotope
t= age of the wine
Ao= initial activity of the wine
A= activity of the at time = t
0.693/12.3 = 2.303/t log (5.5/0.688)
0.693/12.3 = 2.079/t
0.056 = 2.079/t
t= 2.079/0.056
t= 37 years
The wine was produced 37 years ago which means that it was produced in the year 1984
You can tell by how many of the smallest increments there are and the first big number. in this case for number 1 each increment would be worth 1, because it reaches ten and has ten increments. While the one next to it would be worth 2.5, because 10 divided by four equals 2.5. Hope this helps!
