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when 70.0 grams of mno2 reacted with 128.0 grams of hcl, the reaction resulted in a 62.7% yield of chlorine gas. what is the actual yield of chlorine gas in grams? Mno2 + HCI —> MnCl2 + H2O + Cl2​

Sagot :

jbferrado

Answer:

35.7 g

Explanation:

yield(%) = actual yield/theoretical yield x 100

We can calculate the theoretical yield by considering the balanced chemical equation:

MnO₂ + 4HCI —> MnCl₂ + 2H₂O + Cl₂

According to the equation, 1 mol of MnO₂ reacts with 4 moles of HCl to produce 1 mol of Cl₂. So, we can write the following mole ratios:

1 mol MnO₂/4 mol HCl or 4 mol HCl/1 mol MnO₂

1 mol MnO₂/1 mol Cl₂

4 mol HCl/1 mol Cl₂

As we have the amounts of reactants in grams, we have to convert from moles to mass by using the molecular weight (MW) of each compound:

MW(MnO₂) = 54.9 g/mol Mn + (16 g/mol x 2 O) = 86.9 g/mol

1 mol MnO₂ x 86.9 g/mol = 86.9 g MnO₂

MW(HCl) = 1 g/mol H + 35.4 g/mol Cl = 36.4 g/mol

4 moles HCl x 36.4 g/mol = 145.6 g HCl

MW(Cl₂) = 2 x 35.4 g/mol Cl = 70.8 g/mol

1 mol Cl₂ x 70.8 g/mol = 70.8 g Cl₂

Now, we have to figure out which is the limiting reactant. For this, we use the stoichiometric ratio: 145.6 g HCl/86.9 g MnO₂. We multiply the actual amount of MnO₂ by the stoichiometric ratio:

70.0 g MnO₂ x 145.6 g HCl/86.9 g MnO₂ = 117.3 g HCl  < 128.0 g HCl

We need 117.3 grams of HCl to completely react with 70.0 grams of MnO₂, and we have 128 grams of HCl. So, the reactant in excess is HCl, and the limiting reactant is MnO₂.

With the limiting reactant, we calculate the theoretical yield of Cl₂. We use the stoichiometric ratio 70.8 g Cl₂/86.9 g MnO₂:

70.0 g MnO₂ x 70.8 g Cl₂/86.9 g MnO₂ = 57 g Cl₂

Finally, we calculate the actual yield of chlorine gas (Cl₂), by using the first equation:

yield(%) = actual yield/theoretical yield x 100

⇒ actual yield = theoretical yield x yield(%)/100

                       = 57 g x 62.7%/100

                       = 35.7 g

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