Answer to Question #163021 in General Chemistry for Sage Topaz

Question

2.5 g of Al was made to react with 150 mL 0.25 M HCl. What was the limiting reactant? How many g AlCl3 was formed? How many L of H was collected if the reaction took place at 35 deg C and 720 mm Hg? Make the appropriate chemical reaction. balance the chemical equation first. 

Answer 1

The balanced chemical equation is-

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

Mass of Al= 2.5g

Moles of Al = $\dfrac{2.5}{2.7}=0.09259$

Volume of $HCl=150ml$

Molarity of $HCl=0.25M$

No. of moles of $HCl= \dfrac{(150\times 0.25)}{1000} =0.0375$

Here the limiting reagent is Al

So, the number of moles of $AlCl_3=\dfrac{2}{2}\times 0.0925=0.0925$

Mass of $AlCl_3= Moles \times \text{Molar mass of } AlCl_3=0.0925\times133.5=12.41g$

Also, Pressure $P = 720mmHg$ = 1atm

Temprature, $T= 35^{\circ} = 35+273=308K$

Moles of $H_2 = \dfrac{(3\times 0.093)}{2}=\dfrac{0.279}{2}=0.1395$

Using the equation $PV=nRT$

$1\times V=6.1395\times 8.314\times 306$

$V=\dfrac{3957.21}{1000}=0.35721=0.36L$

Hence, the volume of $H_2 \text{ is } 0.36L$