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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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How many grams of NO are produced from 15.8 g of H2O
Yuki888 [10]

Answer:

26.33 g of NO.

Explanation:

We'll begin by writing the balanced equation to produce NO from H2O. This is given below:

5N2 + 6H2O —> 4NH3 + 6NO

Next, we shall determine the mass of H2O that reacted and the mass of NO produced from the balanced equation. This is illustrated below:

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 6 x 18 = 108 g

Molar mass of NO = 14 + 16 = 30 g/mol

Mass of NO from the balanced equation = 6 x 30 = 180 g

From the balanced equation above,

108 g of H2O reacted to produce 180 g of NO.

Finally, we shall determine the mass of NO produced by the reaction of 15.8 g of H2O.

This is illustrated below:

From the balanced equation above,

108 g of H2O reacted to produce 180 g of NO.

Therefore, 15.8 g of H2O will react to produce = (15.8 x 180)/108 = 26.33 g of NO.

Therefore, 26.33 g of NO were obtained from the reaction.

3 0
3 years ago
Gas expands as it heats up. We call this phenomena...
ra1l [238]

I believe the answer is thermal expansion.

7 0
3 years ago
Read 2 more answers
How many moles of mg3n2 would be produced when 180.0 g of mg reacts with 85.0 g of n2?
aalyn [17]
2.47 moles are produce if 85g of N2 react with 180g of Mg
7 0
4 years ago
In a hydrogen fuel cell, hydrogen gas and oxygen gas are combined to form water. Write the balanced chemical equation describing
xxTIMURxx [149]

<u>Answer:</u> The chemical reaction is given below.

<u>Explanation:</u>

A fuel cell is defined as the electrochemical cell which converts the chemical energy of a fuel (often used hydrogen) and an oxidizing agent (often used oxygen) into electrical energy via a pair of redox reactions.

The reactions which occur in hydrogen-oxygen fuel cell are:

At cathode:  H_2+2OH^-\rightarrow 2H_2O+2e^-

At anode:  \frac{1}{2}O_2+H_2O+2e^-\rightarrow 2OH^-

Net reaction:  H_2+\frac{1}{2}O_2\rightarrow H_2O

Thus, the chemical reaction is given above.

5 0
4 years ago
Read 2 more answers
5. In an oxyacetylene welding torch, acetylene (C2H2) burns in pure oxygen with a very hot flame. The reaction is: 2 C2H2 + 5 O2
gulaghasi [49]

11.375 g ethylene is consumed if 35.0 grams of oxygen is allowed to react and 7.875 ml of water will be produced when the 35.0 grams of oxygen is allowed to react.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 ×10^{23} of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

According to the chemical equation for the combustion of ethylene, 5

moles of oxygen are required to react with 2 moles of ethylene, hence

m_{_C_2_H_2}  = 35gX \frac{1 mol}{(2 \;X\; 16)g } X \frac{2 mol}{(5\;mol )} X \frac{26g}{1 mol }

m_{_C_2_H_2} = 11.375 g

Similarly, 5 moles of oxygen are required as a reactant to produce 2 moles of water, thus

m_{_H_2_O} = 35gX \frac{1 mol}{(2 \;X\; 16)g } X \frac{2 mol}{(5\;mol )} X \frac{18g}{1 mol }

m_{_H_2_O} = 7.875 g

Given that the density of water is 1.0g ml^{-1}, therefore

Density = \frac{Mass \;of \;water}{Volume \;of \;water}

Volume \;of \;water = \frac{7.875}{1.0 g ml^{-1}}

The volume of water = 7.875 ml

Learn more about moles here:

brainly.com/question/8455949

#SPJ1

5 0
2 years ago
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