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3 years ago

5. Based on the following reaction: _Fea(SO4)3 +_ KOH > _K2SO4 + _Fe(OH)

Chemistry

2 answers:

Verizon [17]3 years ago

6 0

Answer:

a. Fe₂(SO₄)₃ + 6 KOH → 3 K₂SO₄ + 2Fe(OH)₃

b. 2 mol of KOH

c. 8 mol of Iron(II) hydroxide

Explanation:

This is the balanced reaction

Fe₂(SO₄)₃ + 6 KOH → 3 K₂SO₄ + 2Fe(OH)₃

b. Ratio is 3:6 so,

For 3 mol of potassium sulfate, I needed 6 mol of KOH

For 1 mol of potassium sulfate I would need (1 .6 )/3 = 2 mol of KOH

c. 1 mol of iron(III) sulfate, produce 2 mol of Iron(II) hydroxide

4 mol of iron(III) sulfate, will produce (4x2), 8 mol of Iron(II) hydroxide

ad-work [718]3 years ago

5 0

Answer:

a) Fe2(SO4)3 + 6KOH → 3K2SO4 + 2Fe(OH)3

b) 2 moles of KOH

c) 8 moles of Fe(OH)3

Explanation:

Step 1: The balanced equation

Fe2(SO4)3 + 6KOH → 3K2SO4 + 2Fe(OH)3

Step 2: Data given

Molar mass of K2SO4 = 174.26 g/mol

Molar mass of KOH = 56.11 g/mol

Molar mass of Fe(OH)3 = 106.87 g/mol

Molar mass of Fe2(SO4)3 = 399.88 g/mol

How many mole of potassium hydroxide (KOH) would be needed to create 1 mole of K2SO4 ?

Step 3: Calculate moles of potassium hydroxide (KOH)

For 1 mol Fe2(SO4)3 we need 6 moles of KOH to produce 3 moles of K2SO4 and 2 moles of Fe(OH)3

To produce 1 mol of K2SO4 we need 2*1 = 2 moles of KOH

How many mole of Fe(OH)3, would be produced from reacting 4 mole of Fe2(S04)3 ?

Step 4: Calculate moles of Fe(OH)3

For 1 mol Fe2(SO4)3 we need 6 moles of KOH to produce 3 moles of K2SO4 and 2 moles of Fe(OH)3

When 4 moles of Fe2(SO4)3 will be consumed, there will be produced 2*4 = 8 moles of Fe(OH)3

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What is the percent by volume of isopropyl alcohol in a solution made by mixing 120 mL of the alcohol with enough water to make

Elena L [17]

Answer: The percent by volume of isopropyl alcohol in a solution made by mixing 120 mL of the alcohol with enough water to make 350 mL of solution is 25.5%.

Explanation:

Given: Volume of solute = 120 mL

Volume of solvent = 350 mL

Now, total volume of the solution is as follows.

$V_{total} = V_{solute} + V_{solvent}\\= 120 mL + 350 mL\\= 470 mL$

Let us assume that 100 mL of solution is taken and the amount of isopropyl alcohol present in it is as follows.

$\frac{V_{solute}}{V_{total}} \times 100 mL\\\frac{120 mL}{470} \times 100 mL\\= 25.53 mL$

Hence, there is 25.53 mL isopropyl alcohol is present in 100 mL of solution. Therefore, %v/v is calculated as follows.

$Percent (v/v) = \frac{25.53 mL}{100 mL}\\= 25.5%$

Thus, we can conclude that the percent by volume of isopropyl alcohol in a solution made by mixing 120 mL of the alcohol with enough water to make 350 mL of solution is 25.5%.

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3 years ago

When the reaction 3 no(g) → n2o(g) + no2(g) is proceeding under conditions such that 0.015 mol/l of n2o is being formed each sec

netineya [11]

Answer:

1) The rate of the overall reaction = Δ[N₂O]/Δt = 0.015 mol/L.s.

2) The rate of change for NO = - Δ[NO]/Δt = 3 Δ[N₂O]/Δt = 0.045 mol/L.s.

Explanation:

For the reaction:

3NO(g) → N₂O(g) + NO₂(g).

The rate of the reaction = -1/3 Δ[NO]/Δt = Δ[N₂O]/Δt = Δ[NO₂]/Δt.

Given that: Δ[N₂O]/Δt = 0.015 mol/L.s.

1) The rate of the overall reaction is?

The rate of the overall reaction = Δ[N₂O]/Δt = 0.015 mol/L.s.

2) The rate of change for NO is?

The rate of change for NO = - Δ[NO]/Δt.

∵ -1/3 Δ[NO]/Δt = Δ[N₂O]/Δt.

∴ The rate of change for NO = - Δ[NO]/Δt = 3 Δ[N₂O]/Δt = 3(0.015 mol/L.s) = 0.045 mol/L.s.

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3 years ago

How many milliliters of a 1:2000 drug "i" solution and a 7% drug "i" solution mixed would make 120ml of a 3.5% solution of drug

yanalaym [24]

To solve this problem, let us say that:

x = volume of 1:2000 drug "i" solution

y = volume of 7% drug "i" solution

Assuming volume additive, then this forms:

x + y = 120 mL

x = 120 – y ---> 1

1:2000 also refers to 0.0005 concentrations and 7% also refers to 0.07 concentrations. By doing a component balance:

0.0005 x + 0.07 y = 0.035 (120 mL)

0.0005 x + 0.07 y = 4.2

Substituting equation 1 into this derived equation to get an equation in terms of y:

0.0005 (120 – y) + 0.07 y = 4.2

0.06 – 0.0005 y + 0.07 y = 4.2

0.0695 y = 4.14

y = 59.568 mL = 59.57 mL

From equation 1, x would be:

x = 120 - 59.57

x = 60.43 mL

Answers:

59.57 mL of 1:2000 drug "i" solution

60.43 mL of 7% drug "i" solution

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3 years ago

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Anni [7]

Chemistry is science

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3 years ago

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hydrogen and oxygene react to form water l. How many molecules of O2 are required to produce 0.6 of H20

user100 [1]

Answer:

It is required 0.3 molecules of O2 to produce 0.6 molecules of H2O

Explanation:

The water´s formation reaction is: 2H2 + O2 --> 2H2O

This reaction shows the molecular O2:H2O ratio: 1:2.

Then, if we want to know how many molecules of O2 are required to produce 0.6 of H20, it is necessary calculate as it showed next:

0.6 molecules H2O * (1 molecula O2 / 2 molecules H2O)= 0.3 molecule O2

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3 years ago