Answer:
a. Oxygen is the limiting reagent. 
b. 
%
Explanation:
Hello,
a. Limiting reagent and sulfur trioxide's theoretical yield.
At first, we must compute the involved moles for both sulfur dioxide's and oxygen's as follows, considering the volumes in liters and the pressure in atm of 50.0mmHg*1atm/760mmHg=0.0658atm:
Afterwards, by considering the properly balanced chemical reaction:
We compute the oxygen's moles that completely reacts with the previously computed 
moles of 
as follows:
That result let us know that the oxygen is the limiting reagent since just 
moles are available in comparison with the 
moles that completely would react with moles of .
Now, to compute the theoretical yield of sulfur trioxide, we apply the following stoichiometric relationship:
b. Percent yield.
At first, we must compute the collected (real) moles of sulfur trioxide:
Finally, we compute the percent yield:
%
%
%
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Answer:
There are twice the molar quantity of nitrogen atoms in nitrous oxide, i.e.
1×10−2⋅mol
Moles of nitrous oxide
=
0.217
⋅
g
44.013
⋅
g
⋅
m
o
l
−
1
=
5
×
10
−
3
⋅
m
o
l
.
Given the composition of
N
2
O
, there are this
2
×
5
×
10
−
3
⋅
m
o
l
of nitrogen atoms.
Answer:
496 g of Fe₂O₃.
Explanation:
The balanced equation for the reaction is given below:
4Fe + 3O₂ —> 2Fe₂O₃
From the balanced equation above,
4 moles of Fe reacted to produce 2 moles of Fe₂O₃.
Therefore, 6.20 moles of Fe will react to produce = (6.20 × 2)/4 = 3.1 moles of Fe₂O₃
Finally, we shall determine the mass of 3.1 moles of Fe₂O₃. This can be obtained as follow:
Mole of Fe₂O₃ = 3.1 moles
Molar mass of Fe₂O₃ = (56 × 2) + (3×16)
= 112 + 48
= 160 g/mol
Mass of Fe₂O₃ =?
Mass = mole × molar mass
Mass of Fe₂O₃ = 3.1 × 160
Mass of Fe₂O₃ = 496 g
Therefore, 496 g of Fe₂O₃ were produced from the reaction.
Answer is A
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