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Irina18 [472]
3 years ago
8

4 moles of hydrogen react with 5 moles of oxygen to form water, identify the excess reagent?

Chemistry
1 answer:
Fynjy0 [20]3 years ago
5 0

Answer:

Excess Reagent = oxygen

Explanation:

Limiting reagent: The substance that is totally consumed when the reaction is completed.

Excess reagent: The substance left after the limiting reagent is consumed completely

The balanced chemical equation for formation of water is as follow:

2H_{2}+O_{2}\rightarrow 2H_{2}O

This means when 2 moles of hydrogen reacts with 1 mole of oxygen, 2 moles of water is produced.

Hence the ratio in which hydrogen and oxygen gas reacts is 2:1

Now if 2 mole hydrogen require 1 mole of oxygen ,then 4 mole hydrogen need 2 mole of oxygen.

H_{2}:O_{2}=2:1

or

H_{2}:O_{2}=4:2

Here 5 mole of oxygen is reacting but only 2 mole is required .

Oxygen is in excess.

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To determine the heat required in order to decompose a certain amount of a substance, we need information on the heat needed to decompose one mole of the substance. This value are readily available online and other sources. For this reaction, the heat needed is 129 kJ per 2 mol of NaHCO3. We calculate as follows:

129 KJ / 2 mol NaHCO3 (1 mol / 84.01 g ) (25.5 g NaHCO3 ) = 19.58 kJ of heat is needed.
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400°C

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Nitrogen gas reacts with hydrogen gas to produce ammonia. How many liters of hydrogen gas at 95kPa and 15∘C are required to prod
viktelen [127]

Answer:

222.30 L

Explanation:

We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:

Mass of NH₃ = 100 g

Molar mass of NH₃ = 14 + (3×1)

= 14 + 3

= 17 g/mol

Mole of NH₃ =?

Mole = mass /molar mass

Mole of NH₃ = 100 / 17

Mole of NH₃ = 5.88 moles

Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

3 moles of H₂ reacted to produce 2 moles NH₃.

Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e

Xmol of H₂ = (3 × 5.88)/2

Xmol of H₂ = 8.82 moles

Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:

Pressure (P) = 95 KPa

Temperature (T) = 15 °C = 15 + 273 = 288 K

Number of mole of H₂ (n) = 8.82 moles

Gas constant (R) = 8.314 KPa.L/Kmol

Volume (V) =?

PV = nRT

95 × V = 8.82 × 8.314 × 288

95 × V = 21118.89024

Divide both side by 95

V = 21118.89024 / 95

V = 222.30 L

Thus the volume of Hydrogen needed for the reaction is 222.30 L

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