Standard equation would be N2(g)+3H2(g)==>2NH3(g), so through stoichiometry, (4 mol N2)(2mol NH3/1 mol N2), assuming excess H2, would yield 8 moles of NH3.
KE=.5mv^2
M=mass
v=velocity
.5(4)(100)=200
That should be the answer.
A. We can calculate the initial concentrations of each by
the formula:
initial concentration ci = initial volume * initial
concentration / total mixture volume
where,
total mixture volume = 10 mL + 20 mL + 10 mL + 10 mL = 50
mL
ci (acetone) = 10 mL * 4.0 M / 50 mL = 0.8 M
ci (H+) = 20 mL * 1.0 M / 50 mL = 0.4 M (note: there is only 1 H+ per
1 HCl)
ci (I2) = 10 mL * 0.0050 M / 50 mL = 0.001 M
B. The rate of reaction is determined to be complete when
all of I2 is consumed. This is signified by complete disappearance of I2 color
in the solution. The rate therefore is:
rate of reaction = 0.001 M / 120 seconds
rate of reaction = 8.33 x 10^-6 M / s
Answer:
2.61 g of NO will be formed
The limiting reagent is the O₂
Explanation:
The reaction is:
4NH₃ + 5O₂ → 4NO + 6H₂O
We convert the mass of the reactants to moles:
3.25g / 17 g/mol = 0.191 moles of NH₃
3.50g / 32 g/mol =0.109 moles of O₂
Let's determine the limiting reactant by stoichiometry:
4 moles of ammonia react with 5 moles of oxygen
Then, 0.191 moles of ammonia will react with (0.191 . 5) / 4 = 0.238 moles of oxygen. We only have 0.109 moles of O₂ and we need 0.238, so as the oxygen is not enough, this is the limiting reagent
Ratio with NO is 5:4
5 moles of oxygen produce 4 moles of NO
0.109 moles will produce (0.109 . 4)/ 5 = 0.0872 moles of NO
We convert the moles to mass, to get the answer
0.0872 mol . 30g / 1 mol = 2.61 g
