What mass of CH will be made when 10.0 g of H reacts with 5.0 g of C?

Mass of CH  = 5.46 g

Explanation:

Given data:

Mass of H = 10.0 g

Mass of C = 5.0 g

Mass of CH = ?

Solution:

Chemical equation:

C + H     →   CH

Number of moles of H:

Number of moles = mass/molar mass

Number of moles = 10.0g / 1 g/mol

Number of moles =  10 mol

Number of moles of C:

Number of moles = mass/molar mass

Number of moles = 5.0g / 12 g/mol

Number of moles = 0.42 mol

Now we will compare the moles of H and C with CH.

                      C       :     CH

                        1        :       1

                      0.42    :     0.42

                       H        :        CH

                        1         :       1

                     10         :      10

Number of moles of CH produced by C are less it will act as limiting reactant.

Mass of CH:

Mass = number of moles × molar mass

Mass = 0.42 mol × 13 g/mol

Mass = 5.46 g

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