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4 years ago

If you have 20L of nitrogen, and 25L of hydrogen, how many molecules of NH3 can you make? How much excess reactant is left over?

1 answer:

swat324 years ago

4 0

The balanced equation for formation of NH₃ is as follows;
N₂ + 3H₂ --> 2NH₃
Stoichiometry of N₂ to H₂ is 1:3
when gases react, stoichiometry of reactants applies to volumes as well.
Because 1 mol of any substance occupies 22.4 L at STP.
Therefore stoichiometry of volumes of N₂ to H₂ is 1:3
If N₂ is the limiting reactant,
1 L of N₂ reacts with 3 L of H₂
Therefore 20 L of N₂ should react with - 3/1 x 20 = 60 L
Only 25 L of H₂ is present therefore H₂ is the limiting reactant
ratio of H₂ to NH₃ is 3:2
if 3 L of H₂ forms - 2 L of NH₃
then 25 L of H₂ forms - 2/3 x 25 L = 16.67 L of ammonia
if 22.4 L of NH₃ contains 1 mol of NH₃
then 16.67 L of NH₃ consists of - 1/22.4 x 16.67 =0.74 mol oh NH₃
1 mol of NH₃ is made of 6.022 x 10²³ NH₃ molecules
Therefore 0.74 mol of NH₃ made of 6.022 x 10²³ molecules/mol x 0.74 mol
number of NH₃ molecules formed - 4.45 x 10²³ molecules of NH₃

N₂ is in excess
25 L of H₂ reacts with - 25/3 = 8.33 L
however 20 L present initially
Excess reactant left - 20 - 8.33 = 11.67 L of N₂ remaining

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Hope this helped.

7 0

3 years ago

Complete and balance the reaction in acidic solution. equation: ZnS + NO_{3}^{-} -> Zn^{2+} + S + NO ZnS+NO−3⟶Zn2++S+NO Which

kotykmax [81]

Answer:

Balanced reaction: $3ZnS+2NO_{3}^{-}+8H^{+}\rightarrow 3Zn^{2+}+3S+2NO+4H_{2}O$

S is oxidized and N is reduced.

$NO_{3}^{-}$ is the oxidizing agent and ZnS is the reducing agent.

Explanation:

Reaction: $ZnS+NO_{3}^{-}\rightarrow Zn^{2+}+S+NO$

$\Rightarrow$ Oxidation: $ZnS\rightarrow Zn^{2+}+S$

Balance charge: $ZnS-2e^{-}\rightarrow Zn^{2+}+S$ ...............(1)

$\Rightarrow$ Reduction: $NO_{3}^{-}\rightarrow NO$

Balance H and O in acidic medium : $NO_{3}^{-}+4H^{+}\rightarrow NO+2H_{2}O$

Balance charge: $NO_{3}^{-}+4H^{+}+3e^{-}\rightarrow NO+2H_{2}O$ ...............(2)

[ $3\times$ Equation-(1)] + [ $2\times$ Equation-(2)]:

$3ZnS+2NO_{3}^{-}+8H^{+}\rightarrow 3Zn^{2+}+3S+2NO+4H_{2}O$

Oxidation number of S increases from (-2) to (0) for the conversion of ZnS to S. Therefore S is oxidized.

Oxidation number of N decreases from (+5) to (+2) for the conversion of $NO_{3}^{-}$ to NO. Therefore N is reduced.

$NO_{3}^{-}$ consumes electron from ZnS. Therefore $NO_{3}^{-}$ is the oxidizing agent and ZnS is the reducing agent.

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4 years ago

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irakobra [83]

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6 0

4 years ago

Read 2 more answers

Find the percentage composition of each element in the compound having 9.8 g of nitrogen 0.7 g of hydrogen and 33.6 g of oxygen

Rainbow [258]

Answer:

Search Results

Empirical formula of carbon hydrogen and nitrogen

Nitrogen usually forms three bonds one to Carbon and 2 to Hydrogen atoms. Carbon usually forms four bonds one to Nitrogen and 3 to Hydrogen atoms. The empirical formula is CH5N .

Explanation:

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4 years ago

What mass of sodium hydroxide is needed to neutralise 24.5kg of sulfuric acid <br> H2SO4 + N2 = 2NH3

sveticcg [70]

Answer:

23.92 g

Explanation:

Molar mass of H2SO4 = (2×1)+32+(16×4)= 2+32+48= 82g/mol

H2SO4 + 2NaOH ---> Na2SO4 + 2H2O

I mole of H2SO4 = 2 moles of NaOH

24.5/82 = 24.5/82 × 2

= 0.598 moles of NaOH will neutralize

Mass= mole× molar mass

Molar mass of NaOH= 23+16+1 = 40g/mol

Mass= 0.598 × 40 = 23.92g of NaOH

7 0

3 years ago