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Gwar [14]
3 years ago
14

How many grams of aluminum sulfate form when 3.90 grams of aluminum is placed in 13.65 grams of sulfuric acid?

Chemistry
2 answers:
SVETLANKA909090 [29]3 years ago
4 0

Answer:

Mass = 17.12 g

Explanation:

Given data:

Mass of Al = 3.90 g

Mass of H₂SO₄ = 13.65

Mass of aluminium sulfate = ?

Solution:

Chemical equation:

3H₂SO₄ + 2Al   →  Al₂(SO₄)₃ + 3H₂

Now we will calculate the number of moles of each reactant.

Moles of H₂SO₄:

Number of moles = mass/ molar mass

Number of moles = 13.65 g/ 98.079 g/mol

Number of moles = 0.14 mol

Moles of Al:

Number of moles = mass/ molar mass

Number of moles = 3.90 g/ 27 g/mol

Number of moles = 0.14 mol

Now we will compare the moles of aluminium sulfate with sulfuric acid and aluminium.

                           H₂SO₄       :         Al₂(SO₄)₃

                               3            :              1

                           0.14            :          1/3×0.14 = 0.05

                              Al            :           Al₂(SO₄)₃

                               2            :             1

                            0.14           :        1/2×0.14 = 0.07

The number of moles of aluminium sulfate produced by sulfuric acid are less so it will limiting reactant and limit the amount of aluminium sulfate.

Mass of aluminium sulfate:

Mass = number of moles × molar mass

Mass = 0.05 mol × 342.15 g/mol

Mass = 17.12 g

blondinia [14]3 years ago
3 0

17.1g

Explanation:

Mass of aluminium = 3.9g

Mass of sulfuric acid = 13.65g

Unknown:

Mass of aluminium sulfate = ?

Solution:

To solve this problem, we need to:

   Convert the mass to moles

   Find the limiting reagent

The limiting reagent is the one in short supply in the reaction. It is important to delineate this reagent because it determines the extent of the reaction.

Also, it is important to obtain the balanced reaction equation:

    2Al    +     3H₂SO₄     →      Al₂(SO₄)₃    +   3 H₂

Number of moles = \frac{mass}{molar mass}

  molar mass of Al = 27g/mol

                      H₂SO₄ = 2(1) + 32 + 4(16) = 98g/mol

Number of moles of Al = \frac{3.9}{27} = 0.14mole

Number of moles of H₂SO₄ = \frac{13.65}{98} = 0.14mole

   2 mole of Al reacts with 3 mole of sulfuric acid

   0.14 mole of Al will react with \frac{0.14 x 3}{2} =  0.21mole

      but the amount of sulfuric acid given is 0.14 mole this suggest the sulfuric acid is in short supply. It is the limiting reagent.

To find the number of mole of aluminium sulfate;

             3 mole of sulfuric acid will produce 1 mole of the sulfate

             0.14 mole of the sulfuric acid will produce \frac{0.14}{3} = 0.05mole

Mass of the sulfate = number of moles x molar mass

molar mass of  Al₂(SO₄)₃ = 2(27) + 3(32 + 4(16)) = 342g/mole

 Mass = 0.05 x 342 = 17.1g

learn more:

Molar mass brainly.com/question/2861244

#learnwithBrainly

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marusya05 [52]

Answer:

kb = 2,0x10⁻⁵

Explanation:

The ka for HCN is:

HCN ⇄ H⁺ + CN⁻; ka = 4,9x10⁻¹⁰ <em>(1)</em>

The inverse reaction has an equilibrium constant of:

H⁺ + CN⁻ ⇄ HCN k = 1/4,9x10⁻¹⁰ = 2,0x10⁹ <em>(2)</em>

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The sum of (2) + (3) gives:

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<em />

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5 0
3 years ago
Write the equilibrium constant expression for the reaction below.
kodGreya [7K]

Answer:

Kc = [H₂S]² . [CH₄] / [ H₂O]⁴ . [CS₂]

Explanation:

The equilibrium constant indicates the % of the yield reaction and can shows where the reaction is going to be equilibrated.

It works with molar concentrations on the equilibrium and it does not consider the solids compounds

Kc also can be modified by the time of the reaction.

This reaction is:

CS₂ (g) + 4 H₂O(g) ⇌ CH₄ (g) + 2H₂S (g)

Kc = [H₂S]² . [CH₄] / [ H₂O]⁴ . [CS₂]

8 0
2 years ago
Copper has a density of 8.96 g/cm3. If a thin sheet of copper measures 5.5 cm by 5.5 cm and has a mass of 12.96 g, what is its t
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Thickness of cooper sheet is 0.048 cm (Approx.)

Given that;

Density of copper = 8.96 g/cm³

Measures of copper sheet = 5.5 cm × 5.5 cm

Mass of of copper sheet = 12.96 gram

Find:

Thickness of cooper sheet

Computation:

Volume = Mass / Density

Volume of cooper = 12.96 / 8.96

Volume of cooper = 1.4464

Volume of cooper = 1.45 cm³ (Approx.)

Volume of cooper sheet = 5.5 cm × 5.5 cm x Thickness

5.5 cm × 5.5 cm x Thickness = 1.45

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Thickness of cooper sheet = 0.048 cm (Approx.)

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8 0
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tino4ka555 [31]

Answer:

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3 0
3 years ago
Androstenedione, which contains only carbon, hydrogen, and oxygen, is a steroid hormone produced in the adrenal glands and the g
Gennadij [26K]

Answer: The molecular formula for androstenedione is, C_{19}H_{26}O_2

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=5.527g

Mass of H_2O=1.548g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 5.527 g of carbon dioxide, \frac{12}{44}\times 5.527=1.507g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 1.548 g of water, \frac{2}{18}\times 1.548=0.172g of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = (1.893g)-[(1.507g)+(0.172g)]=0.214g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.507g}{12g/mole}=0.126moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.172g}{1g/mole}=0.172moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.214g}{16g/mole}=0.0133moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0133 moles.

For Carbon = \frac{0.126}{0.0133}=9.5

For Hydrogen  = \frac{0.172}{0.0133}=12.9\approx 13

For Oxygen  = \frac{0.0133}{0.0133}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9.5 : 13 : 1

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of C : H : O = 19 : 26 : 2

Thus, the empirical formula for the given compound is C_{19}H_{26}O_2

The empirical formula weight of C_{19}H_{26}O_2 = 19(12) + 26(1) + 2(16) = 286 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}

n=\frac{286.4}{286}=1

Molecular formula = C_{19}H_{26}O_2

Therefore, the molecular formula for androstenedione is, C_{19}H_{26}O_2

3 0
3 years ago
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