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2 years ago

How many grams of NH3 can be produced from 12.0g of H2?

Chemistry

1 answer:

RSB [31]2 years ago

8 0

Answer:

Balanced reaction:

3 H2 (g) + N2 (g) → 2 NH3 (g)

Use stoichiometry to convert g of H2 to g of NH3. The process would be:

g H2 → mol H2 → mol NH3 → g NH3

12.0 g H2 x (1 mol H2 / 2.02 g H2) x (2 mol NH3 / 3 mol H2) x (17.03 g NH3 / 1 mol NH3) = 67.4 g NH3

Explanation: See above

Hope this helps, friend.

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Given the partial equation: VO2++ Mn2+ → VO2++ MnO4−, balance the reaction in acidic solution using the half-reaction method and

bezimeni [28]

Answer:

Mn^2+(aq) + 4H2O(l) + 5[VO2]^+(aq) + 10H^+(aq) ---------->MnO4^-(aq) + 8H^+(aq) + 5[VO]^2+(aq) + 5H2O(l)

Explanation:

Oxidation half equation:

Mn^2+(aq) + 4H2O(l) ------------> MnO4^-(aq) + 8H^+(aq) + 5e

Reduction half equation:

5[VO2]^+(aq) + 10H^+(aq) + 5e --------> 5[VO]^2+(aq) + 5H2O(l)

Overall redox reaction equation:

Mn^2+(aq) + 4H2O(l) + 5[VO2]^+(aq) + 10H^+(aq) ---------->MnO4^-(aq) + 8H^+(aq) + 5[VO]^2+(aq) + 5H2O(l)

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3 years ago

Determine the value of the equilibrium constant, Kgoal, for the reactionN2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=?by making use of th

Black_prince [1.1K]

Answer:

$K_{goal}=1.793*10^{-33}$

Explanation:

N2(g)+O2(g)⇌2NO(g), $K_1 = 4.10*10^{-31}$

N2(g)+2H2(g)⇌N2H4(g), $K_2 = 7.40*10^{-26}$

2H2O(g)⇌2H2(g)+O2(g), $K_3 = 1.06*10^{-10}$

If we add above reaction we will get:

2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g) Eq (1)

Equilibrium constant for Eq (1) is $K_1*K_3*K_3$

Divide Eq (1) by 2, it will become:

N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g) Eq (2)

Equilibrium constant for Eq (2) is $(K_1*K_3*K_3)^{1/2}$

$Equilibrium constant =K_{goal}= (K_1*K_2*K_3)^{1/2}\\K_{goal}= (4.10*10^{-31} *7.40*10^{-26}*1.06*10^{-10})^{1/2}\\K_{goal}=1.793*10^{-33}$

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3 years ago

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denis-greek [22]

Answer: potential energy is stored

Explanation:

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3 years ago

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C is incorrect because it only applies to nuclear fusion.

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