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 $HCl$.

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

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

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 $\frac{6}{2}\times 0.350=0.704$ 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 $AlCl_3$.

$\text{Mass of }AlCl_3=\text{Moles of }AlCl_3\times \text{Molar mass of }AlCl_3$

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

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