second compound
Let molar mass of x is = X
Let molar mass of y is = Y
Moles of x in second compound = Mass / molar mass = 7 / X
Moles of y in second compound = Mass / molar mass = 4.5 / Y
For second compound
7 / X : 4.5/ Y = 1:1
Therefore
X / Y = 7/4.5
Y / X = 4.5/ 7
The mass of x in first compound = 14g
moles of x in first compound = 14/X
Mass of y in first compound = 3
moles of y in first compound = 3 / Y
14 / X : 3/ Y = 14Y / 3X = 14 X 4.5 / 3 X 7 = 3 :1
Thus molar ratio in first compound = moles of x / Moles of y = 3:2
Formula = x3y
There are 3 equations involved in manufacturing Nitric Acid from Ammonia.
First the ammonia is oxidized:
4NH3 + 5O2 = 4NO + 6H2O
Then for the absorption of the nitrogen oxides.
2NO + O2 = N2O4
Lastly, the N2O4 is further oxidized into Nitric acid.
3N2O4 + 2H2O = 4HNO3 + 2NO
Then run stoichiometry through these equations.
The first equation produces roughly 271,722,938 grams of NO
The second equation produces roughly 416,606,944 grams of N2O4
The last equation produces roughly 380,412,294 grams of HNO3 (nitric acid)
Convert the exact number back into tons, and your answer is: 419.332775 tons.
Rounded, I'm going to say that's 419.33 tons.
Hope this helps! :)
Also, it seems that commercially, Nitric Acid is commonly made by bubbling NO2 into water, rather than using ammonia.
Answer:
0.0238M SbCl3, 1.07M H+, 1.14M Cl-
Explanation:
The total volume of the solution is:
4mL + 5.00mL + 12.0mL = 21mL
As the volume of the SbCl3 is 5.00mL, the dilution factor is:
21mL / 5.00mL = 4.2 times
The concentration of SbCl3 is:
0.10M SbCl3 / 4.2 times = 0.0238M SbCl3
The concentration of H+ = [HCl]:
4.5M / 4.2 times = 1.07M H+
The initial concentration of Cl- is:
3 times SbCl3 + HCl = 0.10M*3 + 4.5M =
<em>3 times SbCl3 because 1 mole of SbCl3 contains 3 moles of Cl-</em>
4.8M Cl- / 4.2 times = 1.14M Cl-
<span>Answer: 100 ml
</span>
<span>Explanation:
1) Convert 1.38 g of Fe₂S₃ into number of moles, n
</span>i) Formula: n = mass in grass / molar mass
<span>
ii) molar mass of </span><span>Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol
</span>
iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of <span>Fe₂S₃
</span>
<span>2) Use the percent yield to calculate the theoretical amount:
</span>
<span>65% = 0.65 = actual yield/ theoretical yield =>
</span>theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol <span>Fe₂S₃</span><span>
3) Chemical equation:
</span>
<span> 3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)
4) Stoichiometrical mole ratios:
</span>
<span>3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl
5) Proportionality:
</span>2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃
<span>
=> x = 0.020 mol FeCl₃
6) convert 0.020 mol to volume
</span>
<span>i) Molarity formula: M = n / V
</span>
<span>ii) V = n / M = 0.020 mol / 0.2 M = 0.1 liter = 100 ml
</span>
