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BartSMP [9]
3 years ago
9

Copper(II) sulfide is formed when copper and sulfur are heated together. In this reaction,

Chemistry
1 answer:
storchak [24]3 years ago
3 0

Answer:

The mass of copper(II) sulfide formed is:

= 81.24 g

Explanation:

The Balanced chemical equation for this reaction is :

Cu(s) + S\rightarrow CuS

given mass= 54 g

Molar mass of Cu = 63.55 g/mol

Moles = \frac{given\ mass}{Molar\ mass}

moles=\frac{54}{63.55}

Moles of Cu = 0.8497 mol

Given mass = 42 g

Molar mass of S = 32.06 g/mol

Moles = \frac{given\ mass}{Molar\ mass}

moles=\frac{42}{32.06}

Moles of S = 1.31 mol

Limiting Reagent :<em> The reagent which is present in less amount and consumed in a reactio</em>n

<u><em>First find the limiting reagent :</em></u>

Cu + S\rightarrow CuS

1 mol of Cu require = 1 mol of S

0.8497 mol of Cu should require  = 1 x 0.8497 mol

= 0.8497 mol of S

S present in the reaction Medium = 1.31 mol

S Required  = 0.8497 mol

S is present in excess and <u>Cu is limiting reagent</u>

<u>All Cu is consumed in the reaction</u>

Amount Cu will decide the amount of CuS formed

Cu + S\rightarrow CuS

1 mole of Cu  gives = 1 mole of Copper sulfide

0.8497 mol of Cu =  1 x 0.8497 mole of Copper sulfide

= 0.8497

Molar mass of CuS = 95.611 g/mol

Moles = \frac{given\ mass}{Molar\ mass}

0.8497 = \frac{given\ mass}{95.611}

Mass of CuS = 0.8497 x 95.611

= 81.24 g

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Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03%
nlexa [21]

<u>Answer:</u> The empirical and molecular formula of chrysotile is Mg_3Si_2H_3O_4 and Mg_6Si_4H_6O_{16}

<u>Explanation:</u>

We are given:

Percentage of Mg = 28.03 %

Percentage of Si = 21.60 %

Percentage of H = 1.16 %

Percentage of O = 49.21 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of Mg = 28.03 g

Mass of Si = 21.60 g

Mass of H = 1.16 g

Mass of O = 49.21 g

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

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Magnesium = \frac{\text{Given mass of Magnesium}}{\text{Molar mass of Magnesium}}=\frac{28.03g}{24g/mole}=1.17moles

Moles of Silicon = \frac{\text{Given mass of Silicon}}{\text{Molar mass of Silicon}}=\frac{21.06g}{28g/mole}=0.752moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.16g}{1g/mole}=1.16moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{49.21g}{16g/mole}=3.07moles

  • <u>Step 2:</u> 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 0.752 moles.

For Magnesium = \frac{1.17}{0.752}=1.5

For Silicon = \frac{0.752}{0.752}=1

For Hydrogen = \frac{1.16}{0.752}=1.5

For Oxygen = \frac{3.07}{0.485}=4.08\approx 4

To convert the mole ratios into whole numbers, we multiply individual mole ratios by 2

Mole ratio of Magnesium = (2 × 1.5) = 3

Mole ratio of Silicon = (2 × 1) = 2

Mole ratio of Hydrogen = (2 × 1.5) = 3

Mole ratio of Oxygen = (2 × 4) = 8

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of Mg : Si : H : O = 3 : 2 : 3 : 8

The empirical formula for the given compound is Mg_3Si_2H_3O_8

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 520.8 g/mol

Mass of empirical formula = [(24 × 3) + (28 × 2) + (1 × 3) + (16 × 8)] = 259 g/mol

Putting values in above equation, we get:

n=\frac{520.8g/mol}{259g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

Mg_{(3\times 2)}Si_{(2\times 2)}H_{(3\times 2)}O_{(8\times 2)}=Mg_6Si_4H_6O_{16}

Hence, the empirical and molecular formula of chrysotile is Mg_3Si_2H_3O_4 and Mg_6Si_4H_6O_{16}

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