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3 years ago

For the following reaction, 4.77 grams of carbon (graphite) are allowed to react with 16.4 grams of oxygen gas.

Chemistry

1 answer:

Katena32 [7]3 years ago

3 0

Answer:

1. 17.5 g of CO₂

2. The limiting reactant is carbon (graphite), and its formula is C(graphite)

3. 3.7 g of O₂

Explanation:

First, we have to write the chemical equation for the reaction. For this, we have to know the chemical formula of each reactant and product:

Reactants: carbon(graphite) ⇒ C(graphite) ; oxygen gas ⇒ O₂(g)
Products: carbon dioxide ⇒ CO₂(g)

Thus, we write the chemical equation:

C(graphite) + O₂(g) → CO₂(g)

The equation is already balanced because it has the same number of C and O atoms on both sides. Thus, we can see that 1 mol of C(graphite) reacts with 1 mol of O₂ and produce 1 mol of CO₂ (mole-to-mole reaction).

Now we convert the grams of reactants to moles by using the molecular weight (Mw) of each compound:

Mw(C) = 12 g/mol

moles of C(graphite) = 4.77g/(12 g/mol) = 0.3975 mol

Mw(O₂) = 16 g/mol x 2 = 32 g/mol

moles of O₂ = 16.4 g/(32 g/mol) = 0.5125 mol

Now, we can compare the stoichiometric ratio (given by the moles of reactants in the equation) with the actual ratio (given by the mass of reactants we have):

stoichiometric ratio ⇒ 1 mol C(graphite)/mol O₂

actual ratio ⇒ 0.3975 mol C(graphite)/0.5125 mol O₂

We can see that we need 0.3975 moles of O₂ to react with C(graphite) and we have more moles (0.5125 mol) so the excess reactant is O₂. Thus, the limiting reactant is C(graphite).

The amount of product (CO₂) that is formed is calculated from the amount of limiting reactant. We can see in the chemical equation that 1 mol of CO₂ is produced from 1 mol of C(graphite) ⇒ stoichiometric ratio = 1 mol CO₂/mol C(graphite).

Thus, we multiply the moles of C(graphite) we have by the stoichiometric ratio to calculate the moles of CO₂ produced:

moles of CO₂ = 0.3975 mol C(graphite) x 1 mol CO₂/mol C(graphite) = 0.3975 mol CO₂

Now, we convert the moles of CO₂ to mass by using the Mw:

Mw(CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol

mass of CO₂ = 0.3975 mol CO₂ x 44 g/mol CO₂ = 17.5 g

Therefore, the maximum amount of carbon dioxide (CO₂) formed is 17.5 g.

Since this is a mole-to-mole reaction, the moles of excess reactant that remains after the reaction is complete is calculated as the difference between the moles of excess reactant and limiting reactant:

remaining moles of O₂ = 0.5125 mol - 0.3975 mol = 0.115 mol O₂

Finally, we convert the moles of O₂ to mass with the Mw (32 g/mol) :

mass of O₂ = 0.115 mol O₂ x 32 g/mol = 3.68 g

Therefore, the mass of the excess reagent that remains after the reaction is complete is 3.7 g.

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3 years ago

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In this, the $CH_3COOH$ is a weak acid so, it not completely dissociated.

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B= B° e^-kt.

Therefore, e^-kt = B/B°.

-kt = ln B/B°.

kt= ln B°/B.

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So, we have; time taken,t = ln( 0.860/.310)/0.500.

==> ln 2.77/0.500.

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3 years ago