Answer: The mass of hydrogen formed when 26.98 g of aluminum reacts with excess hydrochloric acid according to the given balanced equation is 3.03 g.
Explanation:
The given balanced reaction equation is as follows.
[tex]2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}[/tex]
Here, the mole ration of Al and hydrogen produced is 2 : 3
As mass of aluminum is given as 26.98 g. So, moles of aluminum (molar mass = 26.98 g/mol) is as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{26.98 g}{26.98 g/mol}\\= 1 mol[/tex]
So, when 1 mole of Al reacted then 1.5 moles of hydrogen is produced as per the given mole ratio.
Therefore, mass of hydrogen formed is calculated as follows.
[tex]mass = moles \times molar mass\\= 1.5 mol \times 2.02 g/mol\\= 3.03 g[/tex]
Thus, we can conclude that the mass of hydrogen formed when 26.98 g of aluminum reacts with excess hydrochloric acid according to the given balanced equation is 3.03 g.
