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Vesna [10]
3 years ago
10

Ammonia, nh3, for fertilizer is made by causing hydrogen and nitrogen to react at high temperature and pressure. How many moles

of ammonia can be made from 0.15 moles of nitrogen gas? _h2+_02 to _nh3
Chemistry
1 answer:
charle [14.2K]3 years ago
7 0

Answer:

0.30molNH_3

Explanation:

Hello there!

In this case, since the reaction for the formation of ammonia is:

3H_2+N_2\rightarrow 2NH_3

We can evidence the 1:2 mole ratio of nitrogen gas to ammonia; therefore, the appropriate stoichiometric setup for the calculation of the moles of the latter turns out to be:

0.15molN_2*\frac{2molNH_3}{1molN_2}

And the result is:

0.30molNH_3

Best regards!

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Which reactions are oxidation-reduction reactions? check all that apply. fes(s) 2hcl(aq) → h2s(g) fecl2(g) agno3(aq) nacl(aq)
Dafna1 [17]

oxidation-reduction reactions are -

  1. 2C3H6(g) + 9O2(g) → 6CO2(g) + 6H2O(g)
  2. Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

For reaction,

  1. 2C3H6(g) + 9O2(g) → 6CO2(g) + 6H2O(g)

<u>On reactant side</u>:

Oxidation state of Carbon = +2

Oxidation state of Oxygen = 0

<u>On product side:</u>

Oxidation state of Carbon = +4

Oxidation state of Oxygen = -2

Here, carbon's oxidation state is rising from +2 to +4. As a result, it is oxidizing and the oxygen's oxidation state is decreasing from 0 to -2. As a result, it is decreasing.

For reaction,

                Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

<u>When reacting:</u>

Iron's oxidation state is +3.

Carbon's oxidation state is +2.

<u>On product side:</u>

Iron's oxidation state is zero.

Carbon's oxidation state is +4.

Here, carbon's oxidation state is rising from +2 to +4. As a result, it is being oxidized and the iron's oxidation state is changing from +3 to 0. As a result, it is decreasing.

To learn more about oxidation-reduction from given link

brainly.com/question/5794822

#SPJ4

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Answer:

=2.0x10^4mg

Explanation:

Hello there!

In this case, when performing units conversions involving two proportional factors we need to make sure we first convert to the base unit and then to the target one; thus, since 1 kg = 1000 g and 1 g = 1000 mg, we set up the following expression:

=0.020kg*\frac{1000g}{1kg} *\frac{1000mg}{1g} \\\\=2.0x10^4mg

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