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Vika [28.1K]
3 years ago
13

Using the equation below, if you have 4.3 mol of nitrogen tribromide and

Chemistry
1 answer:
ankoles [38]3 years ago
8 0

Answer:

sodium hydroxide is the limiting reactant

Explanation:

The first step is usually to put down the balanced reaction equation. This is the first thing to do when solving any problem related to stoichiometry. The balanced reaction equation serves as a guide during the solution.

2NBr3 + 3NaOH = N2 + 3NaBr + 3HOBr

Let us pick nitrogen gas as our product of interest. Any of the reactants that gives a lower number of moles of nitrogen gas is the limiting reactant.

For nitrogen tribromide

From the balanced reaction equation;

2 moles of nitrogen tribromide yields 1 mole of nitrogen gas

4.3 moles of nitrogen tribromide will yield 4.3 ×1/ 2 = 2.15 moles of nitrogen gas

For sodium hydroxide;

3 moles of sodium hydroxide yields 1 mole of nitrogen gas

5.9 moles of sodium hydroxide yields 5.9 × 1/ 3= 1.97 moles of nitrogen gas

Therefore, sodium hydroxide is the limiting reactant.

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Plz help I’ll give brainliest!!<br> Describe the temperate coniferous forest?<br> Explain
andriy [413]

Answer:

Temperate coniferous forest is a terrestrial biome defined by the World Wide Fund for Nature. This is because Temperate coniferous forests are found predominantly in areas with warm summers and cool winters, and vary in their kinds of plant life.

Explanation:

5 0
2 years ago
Write some example of misicible and immisicible liquid​
Eduardwww [97]

Explanation:

miscible liquids: water and alcohol

immiscible liquids: water and oil

4 0
3 years ago
Determine the number of moles in 3.13 × 10²³ formula units of MgCl₂.
love history [14]

Answer:

<h2>0.52 moles </h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{3.13 \times  {10}^{23} }{6.02 \times  {10}^{23} }  =  \frac{3.13}{6.02}  \\  = 0.519933

We have the final answer as

<h3>0.52 moles</h3>

Hope this helps you

3 0
3 years ago
A student working in the laboratory produces 6.81 grams of calcium oxide, CaO, from 20.7 grams of calcium
xz_007 [3.2K]

Answer:

A. Theoretical yield of CaO is 11.59 g

B. Percentage yield of CaO = 58.76%

Explanation:

The following data were obtained from the question:

Mass of CaCO₃ = 20.7 g

Actual yield of CaO = 6.81 g

Theoretical yield of CaO =?

Percentage yield of CaO =?

The equation for the reaction is given below:

CaCO₃ —> CaO + CO₂

Next, we shall determine the mass of CaCO₃ that decomposed and the mass of CaO produced from the balanced equation. This can be obtained as follow:

Molar mass of CaCO₃ = 40 + 12 + (3×16)

= 40 + 12 + 48

= 100 g/mol

Mass of CaCO₃ from the balanced equation = 1 × 100 = 100 g

Molar mass of CaO = 40 + 16 = 56 g/mol

Mass of CaO from the balanced equation = 1 × 56 = 56 g

SUMMARY:

From the balanced equation above,

100 g of CaCO₃ decomposed to produce 56 g of CaO.

A. Determination of the theoretical yield of CaO.

From the balanced equation above,

100 g of CaCO₃ decomposed to produce 56 g of CaO.

Therefore, 20.7 g of CaCO₃ will decompose to produce =

(20.7 × 56)/100 = 11.59 g of CaO.

Thus, the theoretical yield of CaO is 11.59 g

B. Determination of the percentage yield.

Actual yield of CaO = 6.81 g

Theoretical yield of CaO = 11.59 g

Percentage yield of CaO =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 6.81/11.59 × 100

Percentage yield of CaO = 58.76%

4 0
3 years ago
The chemical equation shown represents photosynthesis. Carbon dioxide plus A plus light with a right-pointing arrow towards B pl
bekas [8.4K]

Answer:

2

Explanation:

8 0
4 years ago
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