Question

20. If 4.95 g of ethylene (C2H4) are combusted with 3.25g of ongen. Hint
a. What is the limiting reagent?
b. How many grams of CO2 are formed?
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Answer 1

Answer:

limiting reagent is oxygen.

amount of co2 formed is 2.97 grams.

thank u

Answer 2

For (a): The limiting reagent is oxygen gas.

For (b): The amount of carbon dioxide produced is 2.99 grams

Explanation:

• For a:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

For ethylene:

Given mass of ethylene = 4.95 g

Molar mass of ethylene = 28 g/mol

Putting values in equation 1, we get:

$\text{Moles of ethylene}=\frac{4.95g}{28g/mol}=0.177mol$

For oxygen gas:

Given mass of oxygen gas = 3.25 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{3.25g}{32g/mol}=0.102mol$

The chemical equation for the reaction of ethylene and oxygen gas follows:

$C_2H_4+3O_2\rightarrow 2CO_2+2H_2O$

By Stoichiometry of the reaction:

3 moles of oxygen gas reacts with 1 mole of ethylene gas

So, 0.102 moles of oxygen gas will react with = $\frac{1}{3}\times 0.102=0.034mol$ of ethylene

As, given amount of ethylene gas is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of oxygen gas produces 2 moles of carbon dioxide

So, 0.102 moles of oxygen gas will produce = $\frac{2}{3}\times 0.102=0.068moles$ of carbon dioxide

• For b:

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.068 moles

Putting values in equation 1, we get:

$0.068mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.068mol\times 44g/mol)=2.99g$

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