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Sever21 [200]
3 years ago
8

If 2.5 moles of nitrogen (N2) react with hydrogen, how many moles of NH3 are produced​

Chemistry
2 answers:
Sedaia [141]3 years ago
7 0
Use the mole ratios of each reactant to work out the total moles of the product
Fynjy0 [20]3 years ago
5 0

Ammonia is produced by the reaction of hydrogen and nitrogen. N2(g)+3H2(g)→2NH3(g)ammonia How many moles of N2 reacted if 0.65mole NH3 is produced? How many moles of NH3 are produced when 1.2moles H2 reacts? 2.In the acetylene torch, does this answer your question?

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How many xenon atoms are contained in 1.00 moles of xenon?
joja [24]

Answer:

6.022 x 10²³

Explanation:

The number of atoms =  the number of moles x with the Avogadro's number.

(The Avogadro's number is 6.022 x 10²³ atoms / moles)

number of atoms = 1.00 moles x 6.022 x 10²³ atoms / mole

number of atoms =  6.022 x 10²³ atoms

(There is no need to simplify?) = 6.022 x 10²³

(ps. This is my first time doing this question so im sorry if i got it wrong

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The density of pentanol, C3H20, is 0.8110 g/mL. Calculate the volume occupied by 7.455 moles of pentanol. What is the volume occ
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3 years ago
What is the atomic weight of silver?
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6 0
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Read 2 more answers
Whats the voltage of CuCl2 + Zn -&gt; ZnCl2 + Cu
gtnhenbr [62]

Answer:

Approximately 1.10\; {\rm V} under standard conditions.

Explanation:

Equation for the overall reaction:

{\rm CuCl_{2}}\, (aq) + {\rm Zn}\, (s) \to {\rm ZnCl_{2}} \, (aq) + {\rm Cu}\, (s).

Write down the ionic equation for this reaction:

\begin{aligned}& {\rm Cu^{2+}}\, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Zn}\, (s)\\ & \to {\rm Zn^{2+}} \, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Cu}\, (s)\end{aligned}.

The net ionic equation for this reaction would be:

{\rm Cu^{2+}}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + {\rm Cu}\, (s).

In this reaction:

  • Zinc loses electrons and was oxidized (at the anode): {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}.
  • Copper gains electrons and was reduced (at the cathode): {\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s).

Look up the standard potentials for each half-reaction on a table of standard reduction potentials.

Notice that {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction, {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s), is reduction and is likely on the table.

  • E(\text{anode}) = -0.7618\; {\rm V} for {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s), and
  • E(\text{cathode}) = 0.3419\; {\rm V} for {\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s).

The reduction potential of {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} would be -E(\text{anode}) = -(-0.7618\; {\rm V}) = 0.7618\; {\rm V}, the opposite of the reverse reaction {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s).

The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:

\begin{aligned} E^{\circ} &= E^{\circ}(\text{cathode}) + (-E^{\circ}(\text{anode})) \\ &= 0.3419 - (-0.7618\; {\rm V}) \\ &\approx 1.10\; {\rm V}\end{aligned}.

7 0
2 years ago
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