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

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.