Answer:
30.4 g. NH3
Explanation:
This problem tells us that the hydrogen (H2) is the limiting reactant, as there is "an excess of nitrogen." Using stoichiometry (the relationship between the various species of the equation), we can see that for every 3 moles of H2 consumed, 2 moles of NH3 are produced.
But before we can use that relationship to find the number of grams of ammonia produced, we need to convert the given grams of hydrogen into moles:
5.4 g x [1 mol H2/(1.008x2 g.)] = 2.67857 mol H2 (not using significant figures yet; want to be as accurate as possible)
Now, we can use the relationship between H2 and NH3.
2.67857 mol H2 x (2 mol NH3/3 mol H2) = 1.7857 mol NH3
Now, we have the number of moles of ammonia produced, but the answer asks us for grams. Use the molar mass of ammonia to convert.
1.7857 mol NH3 x 17.034 g. NH3/mol NH3 = 30.4 g. NH3 (used a default # of 3 sig figs)
Finding percent composition is fairly easy. You only need to divide the mass of an element by the total mass of the compound. We can do this one element at a time.
First, let's find the total mass by using the masses of the elements given on the periodic table.
7 x 12.011 (mass of Carbon) = 84.077
5 x 1.008 (mass of Hydrogen) = 5.04
3 x 14.007 (mass of Nitrogen) = 42.021
6 x 15.999 (mass of Oxygen) = 95.994
Add all of those pieces together.
84.077 + 5.04 + 42.021 + 95.994 = 227.132 g/mol is your total. Since we also just found the mass of each individual element, the next step will be very easy.
Carbon: 84.077 / 227.132 = 0.37016 ≈ 37.01 %
Hydrogen: 5.04 / 227.132 = 0.022189 ≈ 2.22 %
Nitrogen: 42.021 / 227.132 = 0.185 ≈ 18.5 %
Oxygen: 95.994 / 227.132 = 0.42263 ≈ 42.26 %
You can check your work by making sure they add up to 100%. The ones I just found add up to 99.99, which is close enough. A small difference (no more than 0.03 in my experience) is just a matter of where you rounded your numbers.
Answer:
.0556 L
Explanation:
First, convert the 1.35 M to 1.35 mol/L in order for the units to correctly cancel out.
Then, multiply (0.0725 moles Na2CO3/1) times (L/ 1.35 mol).
Finally, the answer will be .0556 L.
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1) Compund Ir (x) O(y)
2) Mass of iridium = mass of crucible and iridium - mass of crucible = 39.52 g - 38.26 g = 1.26 g
3) Mass of iridium oxide = mass of crucible and iridium oxide - mass of crucible = 39.73g - 38.26g = 1.47g
4) Mass of oxygen = mass of iridum oxide - mass of iridium = 1.47g - 1.26g = 0.21g
5) Convert grams to moles
moles of iridium = mass of iridium / molar mass of iridium = 1.26 g / 192.17 g/mol = 0.00656 moles
moles of oxygen = mass of oxygen / molar mass of oxygen = 0.21 g / 15.999 g/mol = 0.0131
6) Find the proportion of moles
Divide by the least of the number of moles, i.e. 0.00656
Ir: 0.00656 / 0.00656 = 1
O: 0.0131 / 0.00656 = 2
=> Empirical formula = Ir O2 (where 2 is the superscript for O)
Answer: Ir O2
The answer would be "air, wood". Gases have the highest Kinetic energy and least Potential energy. Liquids have the 2nd highest Kinetic energy and 2nd least potential energy and solids have the least kinetic energy and highest potential energy.
