Answer:
324.554 K
Explanation:
Which can be converted back to °C by subtracting 273
Once again thus answer can be found using the gas law P1/T1 =P2/T2
Remember to convert to Kelvin before doing the calculations by adding 273
Answer:
Percent yield = 57%
Explanation:
Given data:
Mass of nitrogen = 15.0 g
Mass of hydrogen = 15.0 g
Mass of ammonia produced = 10.5 g
Percent yield = ?
Solution:
Chemical equation:
N₂+ 3H₂ → 2NH₃
Number of moles of hydrogen:
Number of moles = mass/molar mass
Number of moles = 15.0 g/ 2 g/mol
Number of moles = 7.5 mol
Number of moles of nitrogen:
Number of moles = mass/molar mass
Number of moles = 15.0 g/ 28 g/mol
Number of moles = 0.54 mol
Now we will compare the moles of ammonia with nitrogen and hydrogen from balance chemical equation.
N₂ : NH₃
1 : 2
0.54 : 2×0.54 = 1.08
H₂ : NH₃
3 : 2
7.5 : 2/3×7.5= 5 mol
Theoretical yield of ammonia:
Mass = number of moles × molar mass
Mass = 1.08 × 17 g/mol
Mass = 18.36 g
Percent yield:
Percent yield = (actual yield / theoretical yield) ×100
Percent yield = (10.5 g/ 18.36 g) ×100
Percent yield = 0.57 ×100
Percent yield = 57%
1 mole ----------- 6.02x10²³ molecules
? moles ---------- 3.4x10²³ molecules
(3.4x10²³) x 1 / 6.02x10²³ =
3.4x10²³ / 6.02x10²³ => 0.564 moles of H2SO4