The balanced equation for the reaction is,P₄ + 10Cl₂ ---> 4PCl₅Stoichiometry the Cl₂ to PCl₅ 10:4The number of Cl₂ moles reacted - 55.0 g/ 71 g/mol = 0.77 molCl₂ is the limiting reactant, lot of product created depends on lot of limiting reactant present.10 mol of Cl₂ produces - 4 mol that PCl₅when 0.77 mol that Cl₂ reaction - 4/10 x 0.77 = 0.308 mol Therefore 0.308 mol that PCl₅ is formed

Answer : The variety of moles that developed can be 0.321 moles.

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Solution : Given,

Mass the = 57.0 g

Molar fixed of = 71 g/mole

First we have to calculate the mole of . Now we need to calculate the moles of

The balanced chemical reaction is, From the reaction, we conclude that

As, 10 mole the reaction to provide 4 mole of

So, 0.803 mole of react to give moles of

Therefore, the variety of moles that created can it is in 0.321 moles.

Hey there!

Given the reaction:

P4 + 10 Cl2 4 PCl5

Molar massive P4 = 124 g/mol

Number of moles P4:

n = fixed of solute / molar mass

n = 24.0 / 124

n = 0.1935 moles of P4

Therefore:

1 mole P4 4 moles PCl5

0.1935 mole P4 moles PCl5

moles PCl5 = 0.1935 * 4

= 0.774 mole of PCl5

Hope the helps!

We have actually the complying with chemical reaction:

P₄ + 10 Cl₂ → 4 PCl₅

Identifying the limiting reactant.

mass the P₄ = 28 g

mass of Cl₂ = 53 g

number of moles = mass / molecular weight

number of mole of P₄ = 28 / 124 = 0.226 moles

number of mole of Cl₂ = 53 / 71 = 0.746 moles

From the reaction we view that 1 mole the P₄ will certainly react through 10 moles of Cl₂ for this reason 0.226 moles of P₄ will certainly react v 2.226 mole of Cl₂, yet we have only 0.746 mole of Cl₂. So the limiting reactant will certainly be Cl₂.

Part B

From the previous calculations we understand that 28 g of P₄ are equal come 0.226 moles moles the P₄.

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Now, learning the chemical reaction, us devise the following reasoning:

if 1 mole the P₄ produces 4 moles of PCl₅

then 0.226 mole of P₄ to produce X mole of PCl₅

X = (0.226 × 4) / 1 = 0.904 mole of PCl₅

Part C

From the vault calculations we know that 53 g that Cl₂ are equal come 0.746 mole moles the Cl₂.