A 28.5 gram piece of iron is added to a graduated cylinder containing 45.5 mL of water. The water in the cylinder rises to the 49.1 mark. Calculate the density of the iron piece.

7.92 .g . mL^-1

Explanation:

Data provided in the question

Water liters = 45.5 mL

Pace of iron = 28.5 gram

Rise in the water in the cylinder = 49.1 mark

So by considering the above information

As we know that

where,

Mass = 28.5 gram

Volume = 49.1 - 45.5

              = 3.6 mL

So, the density is

= 7.92 .g . mL^-1

Basically we use the density formula by dividing the mass with the volume

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