Question

If 72.0 grams of aluminum are reacted with 252 grams of hydrochloric acid, how many grams of aluminum chloride are produced by this reaction? 2al(s) + 6hcl(aq) → 2alcl3(aq) + 3h2(g)

Answer 1

72.0gAl× 2 mol Al × 2 mol alcl3 × (grams alcl3)
------------ ---------------- ------------------
(grams Al) 2 mol Al 2 mol alcl3

same goes for HCL, remember to use mole to mole ratio

Answer 2

Answer : The mass of
`AlCl_3` produced will be, 46.935 grams.

Explanation : Given,

Mass of
`Al` = 9.5 g

Mass of
`HCl` = 130 g

Molar mass of
`Al` = 26.98 g/mole

Molar mass of
`HCl` = 36.5 g/mole

Molar mass of
`AlCl_3` = 133.34 g/mole

First we have to calculate the moles of
`Al` and
`HCl`.

`\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}=(9.5g)/(26.98g/mole)=0.352moles`

`\text{Moles of }HCl=\frac{\text{Mass of }HCl}{\text{Molar mass of }HCl}=(130g)/(36.5g/mole)=3.56moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`2Al+6HCl\rightarrow 2AlCl_3+3H_2`

From the balanced reaction we conclude that

As, 2 moles of
`Al` react with 6 mole of
`HCl`

So, 0.352 moles of
`Al` react with
`(6)/(2)* 0.350=0.704` moles of
`HCl`

From this we conclude that,
`HCl` is an excess reagent because the given moles are greater than the required moles and
`Al` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`AlCl_3`.

As, 2 moles of
`Al` react to give 2 moles of
`AlCl_3`

So, 0.352 moles of
`Al` react to give 0.352 moles of
`AlCl_3`

Now we have to calculate the mass of
`AlCl_3`.

`\text{Mass of }AlCl_3=\text{Moles of }AlCl_3* \text{Molar mass of }AlCl_3`

`\text{Mass of }AlCl_3=(0.352mole)* (133.34g/mole)=46.935g`

Therefore, the mass of
`AlCl_3` produced will be, 46.935 grams.