Ametal object has a mass of 8.37g. when it was in a graduated cylinder containing 20.0 ml of water, the water level rose to 23.1 ml. what is the density and identity of the metal

The density and identity of the metal would be 2.7 g/ml and aluminum respectively

Mathematically, density = mass/volume

Mass of the object = 8.37 g

Volume of the object = 23.1 - 20

                                 = 3.1 ml

Density of object = 8.37/3.1

                               = 2.7 g/ml

Thus, the density of the object is 2.7 g/ml and the object is most likely an aluminum metal.

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Mass of metal is 8.37 g, volume of water in graduated cylinder is 20 mL. After adding the metal, water level increases to 23.1 mL. The volume of metal is equal to the increase in water level.

The increase in water level can be calculated by taking difference of final and initial level of water.

Thus,

Density of metal is defined as mass per unit volume of metal. It can be calculated as follows:

This value corresponds to the density of aluminum.

Therefore, density is 2.7 g/mL and metal is aluminum.

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g