I am going to go with,
0.10 g/mL
0.0700 g/mL
0.0447 g/mL
I don't know if this is the correct answer, but I am 80% sure that it may be.
:) :)
Answer:
It's Effective Collision.
Explanation:
Hope my answer has helped you!
<span>12.4 g
First, calculate the molar masses by looking up the atomic weights of all involved elements.
Atomic weight manganese = 54.938044
Atomic weight oxygen = 15.999
Atomic weight aluminium = 26.981539
Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol
Now determine the number of moles of MnO2 we have
30.0 g / 86.936044 g/mol = 0.345081265 mol
Looking at the balanced equation
3MnO2+4Al→3Mn+2Al2O3
it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So
0.345081265 mol / 3 * 4 = 0.460108353 mol
So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum.
0.460108353 mol * 26.981539 g/mol = 12.41443146 g
Finally, round to 3 significant figures, giving 12.4 g</span>
The percent yield of the reaction : 89.14%
<h3>Further explanation</h3>
Reaction of Ammonia and Oxygen in a lab :
<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>
mass NH₃ = 80 g
mol NH₃ (MW=17 g/mol):
mass O₂ = 120 g
mol O₂(MW=32 g/mol) :
Mol ratio of reactants(to find limiting reatants) :
mol of H₂O based on O₂ as limiting reactants :
mol H₂O :
mass H₂O :
4.5 x 18 g/mol = 81 g
The percent yield :
The awnser would be c i got you fam
