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3 years ago

What type of reaction does each of the following equations represent?

Chemistry

1 answer:

Reptile [31]3 years ago

3 0

Given :

A chemical equation :

$NH_4OH -> NH_3 + H_2O$

To Find :

What is the type of reaction :

A. Single replacement

B. Decomposition

C. Double replacement

Solution :

We know when a single compound turns into two or more compounds or element, this type of reaction is said to be decomposition reaction.

In the given equation $NH_4OH$ decomposes into $NH_ 3$ and $H_2O$ .

Therefore, the given reaction is decomposition reaction.

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If we mix 25 grans if sodium oxide with a large amount of potassium chloride, how many grams of sodium chloride should be produc

amid [387]

Answer:

47.2 g

Explanation:

Data given:

mass of Sodium oxide (Na₂O) = 25 grams

mass of potassium chloride (KCl) = excess

amount of sodium chloride (NaCl) = ?

Solution:

First we look for reaction of sodium oxide with potassium chloride.

Na₂O + 2 KCl -----------> 2 NaCl + K₂O

As we know that potassium chloride is in excess amount and sodium oxide is 25 g so it means sodium oxide is a limiting reactant and amount of sodium chloride depends on the amount of sodium oxide.

So, now we will look for mole mole ration of Na₂O to NaCl.

Na₂O + 2 KCl -----------> 2 NaCl + K₂O

1 mol 2 mol

from above equation we come to know that 1 mole of Na₂O gives 2 moles of NaCl.

Now convert moles to mass

As we Know

Molar mass of Na₂O = 2(23) + 16 = 62 g/mol

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Na₂O + 2 KCl -----------> 2 NaCl + K₂O

1 mol (62 g/mol) 2 mol (58.5 g/mol)

62 g 117 g

So, it means that 62 g of Na₂O produce 117 g of NaCl then how many grams of NaCl will be produce by 25 g of Na₂O

Apply unity formula

62 g of Na₂O ≅ 117 g of NaCl

25 g of Na₂O ≅ X g of NaCl

Do cross multiplication

X g of NaCl = 117 g x 25 g / 62 g

X g of NaCl = 47.2 g

So,

25 g of Na₂O gives 47.2 grams of NaCl.

4 0

4 years ago

Draw the curved arrow mechanism for the conversion of aniline into benzenediazonium ion, and draw the final product that forms a

kogti [31]

Below Mechanism shows the mechanism of formation of diazonium ion. Aniline is treated with Nitrosonium ion which is generated in situ by the reaction of Sodium nitrile and strong acid. The resulting <span>benzenediazonium ion on reaction with CuBr yields Bromobenzene and on treatment with CuCN gives Benzonitrile. Mechanism is as folow,</span>

3 0

4 years ago

A compound has a pka of 7.4. to 100 ml of a 1.0 m solution of this compound at ph 8.0 is added 30 ml of 1.0 m hydrochloric acid.

Monica [59]

Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.

6 0

3 years ago

Read 2 more answers

What is meant by the limiting reactant why is it necessary to identify the limiting reactant when you want to know what?

WARRIOR [948]

You have the reactants who react to make your products.

Reactants --------> Products

The limiting reactant is the reactant that will RUN OUT FIRST and that will establish the maximum amount of product that will be produced.

To make an example, let's look at this equation and at the following question:

2Al + 6HCl → 2AlCl3 + 3H2

When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed?

We know that 2Al + 6HCL are the reactants, right?

These 2 reactants will react to create an amount of a certain product in this case they're asking us for the amount of H2 that will be produced. Therefore, we need to find the REACTANT it can be either AL OR HCL.

But, how do we know which one is the limiting reactant?

To know which one is the limiting reactant we need to know HOW MUCH H2 can Al produce, and also how much H2 can HCL produce. Why?

It's simple, because if you find the one that produces the less amount of H2, you know immediately that's the maximum amount of that product that will be produced.

Let's say you have a sandwich that needs to be made with 2 slices of bread, 3 meats and 1 cheese.

But you have got 4 slices of bread, 9 slices of meat, and 5 slices of cheese.

Well, you could make 2 sandwiches with 4 slices of bread, 3 sandwiches with those 9 slices of meat and 5 sandwiches with the 5 slices of cheese.

But, in reality, you can only make 2 sandwiches because you don't have any more bread to produce more sandwiches. You get it? That is the point of limiting reactants, to find what is the actual amount that can be produced.

Coming back to our equation, we can find the number of moles of H2 produced by each one of the reactants, Al and HCl. I'll find the number of moles quickly to show you what the concept of limiting reactant is.

6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2

13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2

As you can see, 6.5 mol of H2 is the maximum that can be produced by the HCL and is less than the Al, so that maximum amount that you will get in the product H2 is no more than 6.5.

4 0

4 years ago

In an experiment, a student needs 250.0 mL of a 0.100 M copper (II) chloride solution. A stock solution of 2.00 M copper (II) ch

LekaFEV [45]

Answer : The volume of stock solution needed are, 12.5 mL

Explanation :

Formula used :

$M_1V_1=M_2V_2$