Answer : 21.12 g
explanation :
- A limiting reactant is the substance that is totally consumed when the chemical reaction is complete and the reaction cannot continue without it.
- This is a limiting reactant problem because the amount of product (CO2) formed is limited by this substance (either O2 or C3H8).
- We can predict the limiting reactant by calculating number of mole for all reactants;
n (C3H8) = mass/ molar mass = 7/[(12 × 3)+(1 ×8)] = 0.16 mol
n (O2) = mass/ molar mass = 98/(16 × 2) = 3 mol
But we know from this balanced equation that for the reaction to continue, for (n) of propane there must be (5n) of oxygen. Clearly we have more oxygen than required for the reaction to continue as
[ 3 O2 mol > (0.16 propane mol × 5) . Hence, the limiting reactant is propane.
- Using cross multiplication,
0.16 mol propane → 1 mol propane
? → 3 mol CO2
So (n) of CO2 produced = 0.16 × 3 / 1 = 0.48 mol ,
And mass of CO2 produced = n × molar mass = 0.48 × [12 + (16 ×2)] = 21.12 g..
Answer:
B
Explanation:
We are given that ammonia can be produced from hydrogen gas and nitrogen gas according to the equation:
We want to determine the mass of hydrogen gas that must have reacted if 0.575 g of NH₃ was produced.
To do so, we can convert from grams of NH₃ to moles of NH₃, moles of NH₃ to moles of H₂, and moles of H₂ to grams of H₂.
We are given that the molar masses of NH₃ and H₂ are 17.03 g/mol and 2.0158 g/mol, respectively.
From the equation, we can see that two moles of NH₃ is produced from every three moles of H₂.
With the initial value, perform dimensional analysis:
*Assuming 100% efficiency.
Our final answer should have three significant figures. (The first term has three, the second term has four (the one is exact), the third term is exact, and the fourth term has five. Hence, the product should have only three.)
In conclusion, our answer is B.
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Answer:
A) ΔG° = -3,80x10⁵ kJ
B) E° = 2,85V
Explanation:
A) It is possible to answer this problem using the standard ΔG's of formation. For the reaction:
Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s)
The ΔG° of reaction is:
ΔG° = ΔGFe(s) + ΔGMg²⁺(aq) - (ΔGFe²⁺(aq) + ΔGMg(s) <em>(1)</em>
Where:
ΔGFe(s): 0kJ
ΔGMg²⁺(aq): -458,8 kJ
ΔGFe²⁺(aq): -78,9 kJ
ΔGMg(s): 0kJ
Replacing in (1):
ΔG° = 0kJ -458,8kJ - (-78,9kJ + okJ)
<em>ΔG° = -3,80x10² kJ ≡ -3,80x10⁵ kJ</em>
B) For the reaction:
X(s) + 2Y⁺(aq) → X²⁺(aq) + 2Y(s)
ΔG° = ΔH° - (T×ΔS°)
ΔG° = -629000J - (298,15K×-263J/K)
ΔG° = -550587J
As ΔG° = - n×F×E⁰
Where n are electrons involved in the reaction (<em>2mol</em>), F is faraday constant (<em>96485 J/Vmol</em>) And E° is the standard cell potential
Replacing:
-550587J = - 2mol×96485J/Vmol×E⁰
<em>E° = 2,85V</em>
I hope it helps!
