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Natasha_Volkova [10]
2 years ago
5

The reaction N2 + 3 H2 → 2 NH3 is used to produce ammonia. When 450. g of hydrogen was reacted with nitrogen, 1575 g of ammonia

were produced. What is the percent yield of this reaction?
Chemistry
1 answer:
tiny-mole [99]2 years ago
7 0

Answer:

The percent yield of this reaction is 70%

Explanation:

The reaction is: N₂ + 3H₂ → 2NH₃

We only have the mass of H₂, so we assume that N₂ is in excess

We convert the mass to moles, to work with the reaction:

450 g . 1mol / 2 g = 225 moles

Ratio is 2:3. 3 moles of H₂ can produce 2 moles of ammonia

Therefore 225 moles of H₂ will produce (225 .2)/ 3 = 150 moles

This is the 100% yield reaction → We convert the moles of NH₃ to mass

150 mol . 17g /1mol = 2550 g

Percent yield = (Produced yield/Theoretical yield) .100

Percent yield = (1575g/2550g) . 100 = 70%

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RSB [31]

Answer:

2

Explanation:

Mass of water molecule = mass of hydrated salt - mass of anhydrous salt

Mass of water molecule = 5.00 - 4.26 = 0.74g of water molecule.

Number of moles = mass / molarmass

Molar mass of water = 18.015g/mol

No. of moles of water = 0.74 / 18.015 = 0.0411 moles.

Mass of BaCl2 present =?

1 mole of BaCl2 = 208.23 g

X mole of BaCl2 = 4.26 g

X = (4.26 * 1) / 208.23

X = 0.020

0.020 moles is present in 4.26g of BaCl2

Mole ratio between water and BaCl2 =

0.0411 / 0.020 = 2

Therefore 2 molecules of water is present the hydrated salt.

4 0
2 years ago
How many oxygen atoms are present in a 14.0 g sample of Cu(NO3)2
Vikentia [17]
<span>3 x 2 =6 oxygen atoms and multiply 6 by 14.0 and you have an answer. </span>
4 0
2 years ago
An alpha particle has the same composition as a helium nucleus.<br><br> True<br> False
Schach [20]
Your answer would be True
3 0
2 years ago
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In the reaction of nitrogen gas with oxygen gas to produce nitrogen oxide, what is the effect of adding more oxygen gas to the i
Alborosie
<h3>Answer:</h3>

The Equilibrium would shift to produce more NO

<h3>Explanation:</h3>

The reaction is;

N₂(g) + O₂(g) ⇆ 2NO(g)

  • When a reaction is at equilibrium then the forward reaction rate will be equivalent to the reverse reaction rate. Additionally, the concentration of the reactants and products are the same.
  • From Le Chatelier's principle, additional reactants favor the formation of more products while additional products favor the formation of more reactants.
  • For example, when more oxygen is added then more Nitrogen (II) oxide will be formed.
  • Oxygen is a reactant and when increased it favors forward reaction which leads to the formation of more NO which is the product.

3 0
2 years ago
A student is given the question: "what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume? the density of gold is 19.3
IRINA_888 [86]

<em>The mass of a gold bar  = 1.424 x 10⁴ gram</em>

<em></em>

<h3><em>Further Explanation</em></h3>

Density is a quantity derived from the mass and volume

density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than type

The unit of density can be expressed in g / cm3 or kg / m3

Density formula:

\large{\boxed{{\bold{\rho~=~\frac{m}{V}}}}

ρ = density

m = mass

v = volume

A common example is the water density of 1 gr / cm3

The mass itself is often equated with weight, even though it is different. Mass is the amount of matter in the matter while weight is related to the gravitational force

<em>Known variable</em>

volume gold bar 7.379 × 10⁻⁴ m³

a density of 19.3 g/cm³

<em>Asked</em>

the mass of a gold bar

<em>Answer</em>

volume is known to be 7.379 × 10⁻⁴ m³, then we change it first to units of cm³ adjusting to units of density

7.379 × 10⁻⁴ m³ = 7.379 × 10² cm³

then:

mass = volume x density

mass = 7.379 × 10² cm³ x 19.3 g/cm³

mass = 1.424 x 10⁴ gram

<h3><em>Learn more </em></h3>

density acetone

brainly.com/question/4593217

density and molarity of the solution brainly.com/question/8151398

the percent of acetic acid in the vinegar brainly.com/question/3001349

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Keywords: density, mass, volume, a gold bar

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2 years ago
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