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How many grams of PCl5 can be produced with 16.0 g Cl2 and 23.0 g P4

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Answer:

Therefore, approximately 154.14 grams of PCl5 can be produced from 16.0 grams of Cl2 and 23.0 grams of P4.

Step-by-step explanation:

To determine the number of grams of PCl5 that can be produced, we need to identify the limiting reactant between Cl2 and P4. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.

We'll start by calculating the number of moles for each reactant:

Molar mass of Cl2 = 2 * 35.45 g/mol = 70.90 g/mol

Moles of Cl2 = 16.0 g / 70.90 g/mol = 0.225 mol

Molar mass of P4 = 4 * 31.0 g/mol = 124.0 g/mol

Moles of P4 = 23.0 g / 124.0 g/mol = 0.185 mol

According to the balanced chemical equation for the reaction between Cl2 and P4 to produce PCl5:

P4 + 10Cl2 -> 4PCl5

The stoichiometric ratio between P4 and PCl5 is 1:4. This means that for every 1 mole of P4, 4 moles of PCl5 are produced.

Considering the stoichiometry, we can see that the limiting reactant is P4 because it produces fewer moles of product compared to Cl2.

Now, let's calculate the maximum amount of PCl5 that can be produced from the moles of P4:

Molar mass of PCl5 = 5 * 35.45 g/mol + 1 * 30.97 g/mol = 208.45 g/mol

Moles of PCl5 = 0.185 mol P4 * (4 mol PCl5 / 1 mol P4) = 0.74 mol PCl5

Finally, we can calculate the grams of PCl5:

Mass of PCl5 = Moles of PCl5 * Molar mass of PCl5

Mass of PCl5 = 0.74 mol * 208.45 g/mol ≈ 154.14 g

User Michael Kelley
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