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NARA [144]
3 years ago
13

A dark brown binary compound contains oxygen and a metal. It is 13.38% oxygen by mass. Heating it moderately drives off some of

the oxygen and gives a red binary compound that is 9.334% oxygen by mass. Strong heating drives off more oxygen and gives still another binary compound, which is only 7.168% oxygen by mass.
(a) Compute the mass of oxygen that is combined with 1.000 g of the metal in each of these three oxides.
(b) Assume that the empirical formula of the first compound is MO2 (where M represents the metal). Give the empirical formulas of the second and third compounds.
(c) Name the metal.
Chemistry
1 answer:
Leto [7]3 years ago
6 0

Answer:

a) Mass of O in compound A = 32.72 g

Mass of O in compound B =  21.26 g

Mass of O in compound C = 15.94 g

b) Compound A = MO2

Compound B = M3O4

Compound C = MO

c) M = Pb

Explanation:

Step 1: Data given

A binairy compound contains oxygen (O) and metal (M)

⇒ 13.38 % O

⇒ 100 - 13.38 = 86.62 % M

After heating we get another binairy compound

⇒ 9.334 % O

⇒ 100 - 9.334 = 90.666 % M

After heating we get another binairy compound

⇒ 7.168 % O

⇒ 100 - 7.168 = 92.832 % M

The first compound has an empirical formula of MO2

⇒ 1 mol M for 2 moles O

Step 2: Calculate amount of metal and oxygen in each

compound A:   M  = m1 *0.8662    O = m1 *0.1338

compound B:   M  = m2 *0.90666    O = m2 *0.09334

compound C:   M  = m3 *0.92832    O = m3 *0.07168

Step 3: Calculate mass of oxygen with 1.000 grams of M

Compound A: 1.000g * 0.1338 m1gO  / 0.8662m1gMetal = 0.1545

Compound B: 1.000g * 0.09334 m2gO  / 0.90666m2gMetal = 0.1029

Compound C: 1.000g * 0.07168 m3gO  / 0.92832m3gMetal = 0.07721

Step 4:

1 mol MO2 has 1 mol M and 2 moles O

m1 = (mol O * 16)/0.1338   m1 = 239.2 grams

1 mol M = 0.8632*239.2 = 206.48

0.90666m2 = 206.48  ⇒ m2 = 227.74 g

0.92832m3 = 206.48  ⇒ m3 = 222.42 g

Step 5: Calculate mass of O

Mass of O in compound A = 239.2 - 206.48 = 32.72 g

Mass of O in compound B = 227.74 - 206.48 = 21.26 g

Mass of O in compound C = 222.42- 206.48 = 15.94 g

Step 6: Calculate moles

Moles of O in compound A ≈ 2

⇒ MO2

Moles of O in compound B = 21.26 / 16 ≈ 1.33

⇒ M3O4

Moles of O compound C = 15.94 /16 ≈ 1 moles

⇒ MO

Step 7: Calculate molar mass

The mass of 1 mol metal is 206.48 grams  ⇒ molar mass ≈ 206.48 g/mol

The closest metal to this molar mass is lead (Pb)

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QUESTION 18 Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al (s) 3O 2 (g) 2Al 2O
Delicious77 [7]

Answer:

Percent yield = 85.4%

Explanation:

Given data:

Mass of Al = 2.5 g

Mass of oxygen = 2.5 g

Mass of aluminium oxide = 3.5 g

Percent yield = ?

Solution:

Chemical equation:

4Al + 3O₂  → 2Al₂O₃

First of all we will calculate the number of moles of each reactant.

Number of moles of Al:

Number of moles = mass/ molar mass

Number of moles = 2.5 g/ 27 g/mol

Number of moles = 0.09 mol

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 2.5 g/ 32 g/mol

Number of moles = 0.08 mol

Now we will compare the moles of aluminium oxide with oxygen and aluminium.

                         Al            :          Al₂O₃

                           4           :            2

                        0.09        :         2/4×0.09 = 0.045

                          O₂         :          Al₂O₃

                           3           :            2

                        0.08        :         2/4×0.08 = 0.04

The  number of moles of aluminium oxide produced by oxygen are less so it will limiting reactant.

Theoretical yield of aluminium oxide:

Mass = number of moles × molar mass

Mass = 0.04 mol ×  101.96 g/mol

Mass = 4.1 g

Percent yield:

Percent yield = actual yield /theoretical yield × 100

Percent yield = 3.5 g/ 4.1 g × 100

Percent yield = 85.4%

7 0
3 years ago
Calculate the number of sodium ions, perchlorate ions, Cl atoms and O atoms in 17.8 g of sodium perchlorate. Enter your answers
Degger [83]

17.8 g of sodium perchlorate contains 8.73 × 10²² Na⁺ ions, 8.73 × 10²² ClO₄⁻ ions, 8.73 × 10²² Cl atoms and 3.49 × 10²³ O atoms.

First, we will convert 17.8 g of NaClO₄ to moles using its molar mass (122.44 g/mol).

17.8 g \times \frac{1mol}{122.44g} = 0.145 mol

Next, we will convert 0.145 moles to molecules of NaClO₄ using Avogadro's number; there are 6.02 × 10²³ molecules in 1 mole of molecules.

0.145 mol \times \frac{6.02 \times 10^{23}molecules  }{mol} = 8.73 \times 10^{22}molecules

NaClO₄ is a strong electrolyte that dissociates according to the following equation.

NaClO₄ ⇒ Na⁺ + ClO₄⁻

The molar ratio of NaClO₄ to Na⁺ is 1:1. The number of Na⁺ in 8.73 × 10²² molecules of NaClO₄ is:

8.73 \times 10^{22}moleculeNaClO_4 \times \frac{1 ion Na^{+} }{1moleculeNaClO_4} = 8.73 \times 10^{22}ion Na^{+}

The molar ratio of NaClO₄ to ClO₄⁻ is 1:1. The number of ClO₄⁻ in 8.73 × 10²² molecules of NaClO₄ is:

8.73 \times 10^{22}moleculeNaClO_4 \times \frac{1 ion ClO_4^{-} }{1moleculeNaClO_4} = 8.73 \times 10^{22}ion ClO_4^{-}

The molar ratio of ClO₄⁻ to Cl is 1:1. The number of Cl in 8.73 × 10²² ions of ClO₄⁻ is:

8.73 \times 10^{22}ion ClO_4^{-} \times \frac{1 atomCl }{1ion ClO_4^{-}} = 8.73 \times 10^{22}atom Cl

The molar ratio of ClO₄⁻ to O is 1:1. The number of O in 8.73 × 10²² ions of ClO₄⁻ is:

8.73 \times 10^{22}ion ClO_4^{-} \times \frac{4 atomO }{1ion ClO_4^{-}} = 3.49 \times 10^{23}atom O

17.8 g of sodium perchlorate contains 8.73 × 10²² Na⁺ ions, 8.73 × 10²² ClO₄⁻ ions, 8.73 × 10²² Cl atoms and 3.49 × 10²³ O atoms.

You can learn more Avogadro's number here: brainly.com/question/13302703

8 0
2 years ago
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