All categories

2 years ago

N2 + 3H2 - 2NH3

Chemistry

1 answer:

vladimir1956 [14]2 years ago

4 0

The number of mole of nitrogen gas, N₂, needed to produce 150 g of ammonia, NH₃ is 4.41 moles

<h3>How to determine the mole of NH₃ produced </h3>

Mass of NH₃ = 150 g
Molar mass of NH₃ = 14 + (3×1) = 17 g/mol

Mole of NH₃ =?

Mole = mass /molar mass

Mole of NH₃ = 150 / 17

Mole of NH₃ = 8.82 moles

<h3>How to determine the mole of N₂ needed </h3>

Balanced equation

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

2 moles of NH₃ were produced by 1 mole of N₂.

Therefore,

8.82 moles of NH₃ will be produced by = 8.82 / 2 = 4.41 moles of N₂.

Thus, 4.41 moles of N₂ is needed for the reaction.

Learn more about stoichiometry:

brainly.com/question/14735801

You might be interested in

Since there are many organic compounds,how can you distinguish one organic compound from another

iogann1982 [59]

What are organic and inorganic compounds? Organic chemistry is the study of the carbon compounding molecules. Inorganic chemistry, by contrast, is the study of all compounds that do NOT contain carbon compounds.

8 0

2 years ago

Which statement best describes a physical change?

natta225 [31]

Explanation:

A physical change is one which can be reversed.it also does not produce any new substance.

changes in state such as the melting of a solid, the freezing and vaporization of a liquid as well as changes In shape are examples of physical changes

4 0

2 years ago

An atom's size is determined by how far the outermost electrons are from the nucleus. The size of an atom is affected by the siz

Aleks04 [339]

Answer:

Where are atoms please

Explanation:

4 0

3 years ago

Name five solvents other than water

vovikov84 [41]

Answer:

Ethanol

Methanol

Acetone

Tetrachloroethylene

and Hexanes

Explanation:

5 0

3 years ago

If a solution containing 45.101 g of mercury(II) acetate is allowed to react completely with a solution containing 12.026 g of s

AnnyKZ [126]

Answer:

14.533 grams of solid precipitate of mercury(II) dichromate will form.

Explanation:

$Hg(CH_3COO)_2(aq)+Na_2Cr_2O_7(aq)\rightarrow HgCr_2O_7(s)+2CH_3COONa(aq)$

Moles of mercury(II) acetate = $\frac{45.101 g}{318.70 g/mol}=0.14152 mol$

Moles of sodium dichromate = $\frac{12.026 g}{261.97 g/mol}=0.045906 mol$

According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ of mercury(II) acetate

This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.

According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ of mercury(II) dichromate

Mass of 0.045906 moles of mercury(II) dichromate:

0.045906 mol × 316.59 g/mol = 14.533 g

14.533 grams of solid precipitate of mercury(II) dichromate will form.

3 0

3 years ago