Answer : The variety of moles that developed can be 0.321 moles.
You are watching: How many moles of pcl5 can be produced from 60.0 g of cl2 (and excess p4)?
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
Therefore, the variety of moles that created can it is in 0.321 moles.
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
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.
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
From the vault calculations we know that 53 g that Cl₂ are equal come 0.746 mole moles the Cl₂.