Balance the chemical reaction equation
p4(s)+cl2(g)→pcl5(g)

enter the coefficients in order, separated by commas (e. g., 1,2,3).

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1,10,4

the balanced equation is
p4(s)+10cl2(g)→4pcl5(g)

calculations involving a limiting reactant

now consider a situation in which 28.0 g of p4 is added to 53.0 g of cl2, and a chemical reaction occurs. to identify the limiting reactant, you will need to perform two separate calculations:
calculate the number of moles of pcl5 that can be produced from 28.0 g of p4 (and excess cl2).
calculate the number of moles of pcl5 that can be produced from 53.0 g of cl2 (and excess p4).
then, compare the two values. the reactant that produces the smaller amount of product is the limiting reactant.
part b

how many moles of pcl5 can be produced from 28.0 g of p4 (and excess cl2)?

express your answer numerically in moles.

part c

how many moles of pcl5 can be produced from 53.0 g of cl2 (and excess p4)?

We have the following chemical reaction:

P₄ + 10 Cl₂ → 4 PCl₅

Identifying the limiting reactant.

mass of  P₄ = 28 g

mass of Cl₂ = 53 g

number of moles = mass / molecular weight

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

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

From the reaction we see that 1 mole of P₄ will react with 10 moles of Cl₂ so 0.226 moles of P₄ will react with 2.226 moles of Cl₂, but we have only 0.746 moles of Cl₂. So the limiting reactant will be Cl₂.

Part B

From the previous calculations we know that 28 g of P₄ are equal to 0.226 moles moles of P₄.

Now, knowing the chemical reaction, we devise the following reasoning:

if        1 mole of P₄ produces 4 moles of PCl₅

then  0.226 moles of P₄ produces X moles of PCl₅

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

Part C

From the previous calculations we know that 53 g of Cl₂ are equal to 0.746 moles moles of Cl₂.

Now, knowing the chemical reaction, we devise the following reasoning:

if         10 moles of Cl₂ produces 4 moles of PCl₅

then   0.746 moles of Cl₂ produces X moles of PCl₅

X = (0.746 × 4) / 10 = 0.298 moles of PCl₅

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year