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Korolek [52]
4 years ago
10

Does the amount of methanol increase, decrease, or remain the same when an equilibrium mixture of reactants and products is subj

ected to the following changes?
a. the catalyst is removed
b. the temp is increased
c. the volume is decreased
d. helium is added
e. CO is added
Chemistry
1 answer:
denis23 [38]4 years ago
3 0

Answer:

a. Methanol remains the same

b. Methanol decreases

c. Methanol increases

d. Methanol remains the same

e. Methanol increases

Explanation:

Methanol is produced by the reaction of carbon monoxide and hydrogen in the presence of a catalyst as follows; 2H2+CO→CH3OH.

a) The presence or absence of a catalyst makes no difference on the equilibrium position of the system hence the methanol remains constant.

b) The amount of methanol decreases because the equilibrium position shifts towards the left and more reactants are formed since the reaction is exothermic.

c) If the volume is decreased, there will be more methanol in the system because the equilibrium position will shift towards the right hand side.

d) Addition of helium gas has no effect on the equilibrium position since it does not participate in the reaction system.

e) if more CO is added the amount of methanol increases since the equilibrium position will shift towards the right hand side.

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Combustion analysis of 0.600 g of an unknown compound containing carbon, hydrogen, and oxygen produced 1.043 g of CO2 and 0.5670
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Answer : The empirical formula of the compound is C_3H_8O_2

Explanation :

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=1.043g

Mass of H_2O=0.5670g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 1.043 g of carbon dioxide, \frac{12}{44}\times 1.043=0.284g of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.5670 g of water, \frac{2}{18}\times 0.5670=0.063g of hydrogen will be contained.

Mass of oxygen in the compound = (0.600) - (0.284 + 0.063) = 0.253 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.284g}{12g/mole}=0.0237moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.063g}{1g/mole}=0.063moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.253g}{16g/mole}=0.0158moles

Step 2: 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 0.0158 moles.

For Carbon = \frac{0.0237}{0.0158}=1.5

For Hydrogen  = \frac{0.063}{0.0158}=3.98\approx 4

For Oxygen  = \frac{0.0158}{0.0158}=1

The ratio of C : H : O = 1.5 : 4 : 1

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of C : H : O = 3 : 8 : 2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 8 : 2

Hence, the empirical formula for the given compound is C_3H_8O_2

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