Calculate the mass of hydrogen formed when 26.98 g of aluminum reacts with excess hydrochloric acid according to the following balanced chem

Question

Calculate the mass of hydrogen formed when 26.98 g of aluminum reacts with excess hydrochloric acid according to the following balanced chemical equation: 2 Al + 6 HCl → 2 AlCl3 + 3 H2

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Vân Khánh 6 months 2021-07-23T06:03:25+00:00 1 Answers 21 views 0

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    2021-07-23T06:05:22+00:00

    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.

    2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}

    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.

    Moles = \frac{mass}{molar mass}\\= \frac{26.98 g}{26.98 g/mol}\\= 1 mol

    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.

    mass = moles \times molar mass\\= 1.5 mol \times 2.02 g/mol\\= 3.03 g

    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.

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