How many grams of water are needed to produce 28 grams of aluminum carbide (al4c3)? al4c3 + h2o --> 3ch4 + 4 al (oh)3

mass of H₂O = 19.44 g

Note: There is a typing mistake in question, its is Aluminum hydroxide not aluminum carbide, as we can see in the given reaction.

Explanation:

balanced chemical equation

  Al₄C₃ + 12 H₂O → 4 Al(OH)₃ + 3 CH₄

given data

    moles of H₂O = 12 mol

    mas of H₂O = ?

    moles of Al(OH)₃ = 4 mol

    mass of Al(OH)₃ = 28 g

Solution

1st we will find out the mole ratio of water and Aluminum hydroxide from balanced chemical equation

       H₂O : Al(OH)₃

           12 : 4

        12/4 : 4/4

             3 : 1

Now we find out number of moles of Aluminum hydroxide

           moles = mass / molar mass

  moles = 28 g / 78 g/mol

  moles = 0.36 mol

now we find out moles (x) of H₂O needed for 0.36 mol of Aluminum hydroxide

from the balanced chemical equation the mole ratios are:

              3 : 1

              x : 0.36

Cross multiply these ratios

     3 × 0.36 = 1 x

     1.08 = 1 x

     x = 1.08 mol

Now we will find out the mass of H₂O

   mass of H₂O = moles × molar mass

   mass of H₂O = 1.08 × 18

   mass of H₂O = 19.44 g

1 a

2 d

3 a

4 c

explanation: