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Ipatiy [6.2K]
4 years ago
8

Find the volume of 0.120 M hydrochloric acid necessary to react completely with 1.55 g Al(OH)3.

Chemistry
2 answers:
monitta4 years ago
7 0
The balanced chemical reaction is expressed as:

3HCl + Al(OH)3 = AlCl3 + 3H2O

We are given the amount of aluminum hydroxide to be used up in the reaction. This value will be the starting point for the calculations. We calculate as follows:

1.55 g Al(OH)3 ( 1 mol Al(OH)3 / 78 g Al(OH)3 )( 3 mol HCl / 1 mol Al(OH)3 ) = 0.06 mol HCl

Volume = 0.06 mol HCl / 0.120 mol HCl/ L solution = 0.50 L solution
blagie [28]4 years ago
6 0

<u>Answer:</u> The volume of HCl is 55.83 mL

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of aluminium hydroxide = 1.55 g

Molar mass of aluminium hydroxide = 78 g/mol

Putting values in above equation, we get:

\text{Moles of aluminium hydroxide}=\frac{1.55g}{78g/mol}=0.020mol

The chemical reaction for the formation of chromium oxide follows the equation:

Al(OH)_3+3HCl\rightarrow AlCl_3+6H_2O

By Stoichiometry of the reaction:

1 mole of aluminium hydroxide reacts with 3 moles of HCl

So, 0.020 moles of aluminium hydroxide will react with = \frac{1}{3}\times 0.020=0.0067mol of HCl

To calculate the volume of HCl, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

We are given:

Moles of HCl = 0.0067 moles

Molarity of solution = 0.120 M

Putting values in above equation, we get:

0.120=\frac{0.0024\times 1000}{\text{Volume of HCl}}\\\\\text{Volume of HCl}=\frac{0.0024\times 1000}{0.120}=55.83mL

Hence, the volume of HCl is 55.83 mL

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At 46°C a sample of ammonia gas exerts a pressure of 5.3 atm. What is the pressure when the volume of the gas is reduced to one-
snow_lady [41]

Hello there!

As you may know, pressure and volume have an inverse relationship. This means that as pressure increases, volume decreases (and as volume increases, pressure decreases).

In this problem, the volume of the gas is reduced to 0.250 its original value. This means that the pressure should be higher than it was originally.

We can use Boyle's Law to solve this problem, seeing that temperature is kept constant.

P1V1 = P2V2

Let's rearrange this equation to isolate the variable we need, P2.

P1V1

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 V2

That's better! Now that we have that down, let's plug in the numbers we were given.

(5.3atm) * (V1)

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0.25(V1)

We can get rid of V1, as the number would be equal to 1 (anything divided by itself is equal to 1).

5.3atm

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Now, while the number is significantly higher, this is quite reasonable.

Let's plug it in and see if we get what we should:

5.3(V1) = (0.25V1)(21.2)

--------

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5 0
3 years ago
It takes 42.25 ml of 0.0500 M Na2S2O3 solution to completely react with the iodine present in a 150.0 ml iodine solution. How ma
svlad2 [7]

Answer:

There were 0.268 grams I2 present

Explanation:

Step 1: Data given

Volume of Na2SO3 = 42.25 mL = 0.04225 L

Molarity of Na2SO3 = 0.0500 M

Volume of iodine = 150.0 mL

Step 2: The balanced equation

I2 + 2 Na2SO3 → Na2S2O6 + 2 NaI

Step 3: Calculate moles of Na2SO3

Moles Na2SO3 = molarity * volume

Moles Na2SO3 = 0.0500 M * 0.04225 L

Moles Na2SO3 = 0.0021125 moles

Step 4: Calculate moles I2

For 1 mol I2 we need 2 moles Na2SO3

For 0.0021125 moles Na2SO3 we have 0.00105625

Step 5: Calculate mass of I2

Mass I2 = moles * molar mass

Mass I2 = 0.00105625 * 253.8 g/mol

Mass I2 = 0.268 grams

There were 0.268 grams I2 present

3 0
4 years ago
Read 2 more answers
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