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Natali [406]
3 years ago
8

lead sulfide reacts with oxygen gas to produce lead oxide and sulfur dioxide. if 0.500 moles of o2 were consumed using this chem

ical reaction, how many moles of lead oxide would be produced
Chemistry
1 answer:
Digiron [165]3 years ago
7 0
Easy!

All you have to do is look at the formula.

Lead Sulfide is PbS

Lead Oxide is PbO

Sulfur Dioxide is SO₂

Oxygen is O₂

You would need to balance out the equation, so it would be PbS + O₂ --> PbO + SO₂.

We can balance this out with the following.  2PbS + 2O₂ --> 2PbO + SO₂.

So 0.500 moles of oxygen would be the same conversion rate for PbO so it would also be 0.500 moles of lead oxide (PbO)
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The smell of dirty gym socks is caused by the compound caproic acid(contains H, C, O). Combustion of 0.844 g of caproic acid pro
adoni [48]

<u>Answer:</u> The molecular formula for the given organic compound is C_2H_{32}O_4

<u>Explanation:</u>

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=0.784g

Mass of H_2O=1.92g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

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

So, in 0.784 g of carbon dioxide, \frac{12}{44}\times 0.784=0.214g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

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

So, in 1.92 g of water, \frac{2}{18}\times 1.92=0.213g of hydrogen will be contained.

Mass of oxygen in the compound = (0.844) - (0.214 + 0.213) = 0.417 g

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

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

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.214g}{12g/mole}=0.0134moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.213g}{1g/mole}=0.213moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.417g}{16g/mole}=0.0261moles

  • <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 0.0134 moles.

For Carbon = \frac{0.0134}{0.0134}=1

For Hydrogen = \frac{0.213}{0.0134}=15.89\approx 16

For Oxygen = \frac{0.0261}{0.0134}=1.95\approx 2

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

The ratio of C : H : O = 1 : 16 : 2

The empirical formula for the given compound is CH_{16}O_2

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 = 116.2 g/mol

Mass of empirical formula = 60 g/mol

Putting values in above equation, we get:

n=\frac{116.2g/mol}{60g/mol}=1.94\approx 2

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

C_{(1\times 2)}H_{(16\times 2)}O_{(2\times 2)}=C_2H_{32}O_4

Hence, the molecular formula for the given organic compound is C_2H_{32}O_4

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Explanation:

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