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OLEGan [10]
3 years ago
9

How many grams of aluminum sulfide would form if 2.25 liters of a 1.50 molar aluminum chloride solution reacts with 2.00 liters

of a 2.75 molar hydrosulfuric acid (H2S) solution?
2AlCl3 + 3H2S Al2S3 + 6HCl (2 points)
• 18.6 g
• 101 g
• 253 g
• 275 g
Chemistry
2 answers:
kogti [31]3 years ago
8 0

275g is the correct answer

Brums [2.3K]3 years ago
5 0
1) Calculate the number of mols of each reagent, from the molar concentrations.

<span>2.25 liters of a 1.50 molar aluminum chloride solution

M = n / V => n = M*V = 2.5liters*1.50 mol/liter = 3.75 mol

 2.00 liters of a 2.75 molar hydrosulfuric acid (H2S) solution

n = 2.00 liter * 2.75 mol/liter = 5.5 mol

2AlCl3 + 3H2S -> Al2S3 + 6HCl

Theoretical ratio 3 ACl3 / 2 H2S
Actual ratio 3.75 / 5.5 = 0.68 > 2/3 => H2S is the limitan reagent

5.5 mol H2S * 1 mol Al2S3 /  3mol H2S = 1.833 mol Al2S3

Molar mass of Al2S3 = 2* 27 g/mol + 3*32 g/mol = 150 g/mol

Mass of Al2S3 = 150g/mol*1.833mol = 274.95 g

Answer: 275 g
</span>

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What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m
djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the  solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of  sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL  (1cm³  = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³  mol / 0.4327 L = 0.0075  M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x  432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for  volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

8 0
3 years ago
A beaker of water has a volume of 125mL and a density of 1.0g/mL. Calculate the mass of the water.​
Butoxors [25]

Answer:

125 g

Explanation:

D = m/v

1.0 g/mL = m /(125 mL)

125 g = m

6 0
3 years ago
How many chlorine atoms are there in 12.2 g of ccl4? How many chlorine atoms are there in 12.2 of ? 4.77×1022 atoms?
mash [69]

Answer:- 1.91*10^2^3Cl atoms.

Solution:- We have been given the grams of carbon tetrachloride and asked to calculate the number of atoms of chlorine. It is a three step conversion problem. In the first we convert the grams of carbon tetrachloride to moles of it. In second step we convert moles of carbon tetrachloride to moles of chlorine and in the third step we convert the moles of chlorine to atoms of chlorine.

For grams to mole conversion we need the molar mass of the compound. Molar mass of carbon tetrachloride is 153.82 grams per mol. If we look at the formula of carbon tetrachloride then four chlorine are present in it. It means 1 mol of carbon tetrachloride has four moles of chlorine. The calculations are as follows:

12.2gCCl_4(\frac{1molCCl_4}{153.82gCCl_4})(\frac{4molCl}{1molCCl_4})(\frac{6.02*10^2^3Clatoms}{1molCl})

= 1.91*10^2^3Cl atoms

So, there are 1.91*10^2^3Cl atoms in 12.2 grams of CCl_4 .

6 0
3 years ago
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Answer:Yes

Explanation:

3 0
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Vera_Pavlovna [14]
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3 0
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