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vitfil [10]
4 years ago
14

Na reação de óxido de alumínio com ácido sulfúrico forma-se sulfato de alumínio (equação abaixo). Levando em consideração que a

reação tenha um rendimento total, para produzir 6 mols desse sulfato, qual é a quantidade de matéria, em mol, de ácido necessária?
(olhe a imagem por favor)

Chemistry
1 answer:
never [62]4 years ago
7 0

Answer:

18 moles de ácido

Explanation:

A reação de óxido de alumínio com ácido sulfúrico é:

Al₂O₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂O

<em>Onde 1 mol de óxido de alumínio reaciona com 3 moles de ácido sulfúrico produzendo 1 mol de sulfato de alumínio e 3 moles de água.</em>

Agora, se você quizer produzir 6 moles de sulfato de alumínio, e você sabe que 3 moles de ácido formam 1 mol de sulfato, as moles de ácido que você precisa são:

6 moles Al₂(SO₄)₃ × ( 3 moles H₂SO₄ / 1mole Al₂(SO₄)₃) = <em>18 moles de ácido</em>

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How many milliliters of 0.105 M HCl are needed to titrate each of the following solutions to the equivalence point?
kolezko [41]

<u>Answer:</u>

<u>For a:</u> The volume of HCl needed is 4.52 mL

<u>For b:</u> The volume of HCl needed is 25.63 mL

<u>For c:</u> The volume of HCl needed is 43.33 mL

<u>Explanation:</u>

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base

  • <u>For a:</u>

We are given:

n_1=1\\M_1=0.105M\\V_1=?mL\\n_2=1\\M_2=9.50\times 10^{-2}M=0.00950M\\V_2=50.0mL

Putting values in above equation, we get:

1\times 0.105\times V_1=1\times 0.00950\times 50\\\\V_1=\frac{1\times 0.00950\times 50}{1\times 0.105}=4.52mL

Hence, the volume of HCl needed is 4.52 mL

  • <u>For b:</u>

We are given:

n_1=1\\M_1=0.105M\\V_1=?mL\\n_2=1\\M_2=0.117M\\V_2=23.0mL

Putting values in above equation, we get:

1\times 0.105\times V_1=1\times 0.117\times 23\\\\V_1=\frac{1\times 0.117\times 23}{1\times 0.105}=25.63mL

Hence, the volume of HCl needed is 25.63 mL

  • <u>For c:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Given mass of NaOH = 1.40 g

Molar mass of NaOH = 40 g/mol

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of NaOH solution}=\frac{1.40}{40\times 1}\\\\\text{Molarity of NaOH}=0.035M

We are given:

n_1=1\\M_1=0.105M\\V_1=?mL\\n_2=1\\M_2=0.035M\\V_2=130mL

Putting values in above equation, we get:

1\times 0.105\times V_1=1\times 0.035\times 130\\\\V_1=\frac{1\times 0.035\times 130}{1\times 0.105}=43.33mL

Hence, the volume of HCl needed is 43.33 mL

6 0
3 years ago
A container at stp contains 3.46 moles of neon gas. What is the volume of the neon gas?
ziro4ka [17]
3.46 moles x 22.4 L/1 mol neon gas = 77.5 L neon gas
3 0
3 years ago
Force = 2.32 N<br> Acceleration = 0.19 m/s2<br> Mass= ?
pochemuha

m=f/a, so m=2.32/.19

4 0
4 years ago
Calculate the number of moles of aluminum, sulfur, and oxygen atoms in 9.00 moles of aluminum sulfate, Al2(SO4)3.
vazorg [7]

The formula of aluminum sulfate is Al₂(SO₄)₃

Hence, 1 mole of Al₂(SO₄)₃ has 2 moles of Al, 3 moles of S and (4 x 3) or 12 moles of O

9 moles of Al₂(SO₄)₃ x (2 moles of Al/1 mole of Al₂(SO₄)₃) = 18 moles of Al

9 moles of Al₂(SO₄)₃ x (3 moles of S/1 mole of Al₂(SO₄)₃) = 27 moles of S

9 moles of Al₂(SO₄)₃ x (12 moles of O/1 mole of Al₂(SO₄)₃) = 108 moles of O

The number of moles of Al, S, and O atoms are 18, 27 and 108.

3 0
3 years ago
Sodium hypochlorite (Na xCl yO z) is the active ingredient in household bleach. A 250.0 g sample of sodium hypochlorite contains
zimovet [89]

Answer:

Empyrical formula → NaClO

Explanation:

To determine the empirical formula of sodium hypochlorite we need the centesimal composition:

Grams of an element in 100 g of compound.

77.1 g of Na in 250 g of compound

119.1 g of Cl in 250 g of compound

(250g - 119.1g - 77.1g) = 53.8 g of O in 250 g of compound

We make this rules of three:

In 250 g of compound we have 77.1 g of Na, 119.1 g of Cl and 53.8 g of O

In 100 g of compound we must have:

(77.1 . 100) / 250 = 30.84 g of Na

(119.1 . 100) / 250 = 47.64 g of Cl

(53.8 . 100) / 250 = 21.52 g of O

We divide the mass by the molar mass of each element:

30.84 g / 23 g/mol = 1.34 mol of Na

47.64 g / 35.45 g/mol = 1.34 moles of Cl

21.52 g / 16 g/mol = 1.34 mol of O

We divide by the lowest value of obtained mol, but in this case, it's the same number for the three of them.

In conclussion, the empyrical formula for the sodium hypochlorite is: NaClO

3 0
3 years ago
Read 2 more answers
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