A form of phosphorus called red phosphorus is used in match heads. When 0.062 g of red phosphorus burns in air, it forms 0.142 g of phosphorus oxide. Determine the empirical formula of phosphorus oxide.(show work; use labels)

The empirical formula is

Explanation:

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

a) If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of P = 0.062 g

Mass of O= (0.142- 0.062) = 0.08 g

Step 1 : convert given masses into moles

Moles of P=

Moles of O=

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For P =

For O =

The ratio of P: O = 1: 2.5

Converting them into whole number ratios by multiplying with 2:

Hence the empirical formula is

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2