What is the limiting reactant and what is the maximum mass of aluminum hydride that can be formed given 3.00 grams of hydrogen and 26.00 grams of aluminum ?

Now let's balance it! We can see that the Al molecules are balanced but the Hydrogens are not, so we can multiply both molecules by the opposite number:

3H2 + Al -> 2AlH3

Now we have 6 hydrogens on each side, we just add a 2 to balance the aluminum and we have the balanced equation!

3H2 + 2Al -> 2AlH3

Step 2: Convert all values to no. of moles

To do this, we can use the formula:

mass = no. moles x Molar mass

Which we can rearrange to find No. moles = mass/ Molar mass

Hydrogen: 3.00/2.02 = 1.49 moles

Aluminum: 26.00/26.98 = 0.9637 moles

Step 3: Find the Limiting Reagent

Now we choose one (let's use hydrogen) and divide by the co-efficient ratio between the two elements (3H2 for every 2 Al).

1.49 x 2/3 = 0.993 (This is the theoretical amount of Aluminum we need for the amount of hydrogen we have)

Now we compare this number with the actual no. moles of aluminum we have to find the limiting reagent:

0.993 > 0.9637

This means that Aluminum is the limiting reagent. (We are working with 0.9637 moles of Aluminum.)

Step 4: Find amount of product formed

We can use the molar ratio of the balanced equation again to find how much AlH3 is made. The coefficient of Al is the same as the coefficient of AlH3. This means that equal no. if moles are made in the reaction.

This means that 0.9637 moles of Aluminum Hydride is formed in this reaction.

Step 5: Convert back to mass in grams

Now we just need to convert the no. moles back to mass using the same formula:

mass = no. moles x Molar Mass

m = 0.9637 x (26.98 + 3 x 1.01)

m = 0.9637 x 30.01

m = 28.92g (4.s.f.)

Hope this helped!