7. A certain mass of water was heated with 41,840 Joules, raising its
temperature from 22.0 °C to 28.5 °C. Find the mass of water.

mass of water = 1540 g

Explanation:

Given data:

Heat absorbed = 41840 j

Initial temperature = 22.0°C

Final temperature = 28.5 °C

Mass of water = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water = 4.18 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 28.5 °C - 22 °C

ΔT =  6.5°C

Now we will  put the values in formula:

41840 J = m ×4.18 j/g.°C× 6.5°C

41840 J = m × 27.17 j/g

m = 41840 J /27.17 j/g

m = 1540 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