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