A student performed the following reaction in his Chemistry class: Zn + 2HCl => ZnCl2+H2. He started with 32.70 g of Zn metal and 18.23

Question

A student performed the following reaction in his Chemistry class: Zn + 2HCl —> ZnCl2+H2. He started with 32.70 g of Zn metal and 18.23 g HCl and got 60.1 g of of ZnCl2.

(I) How much ZnC12 should he have produced?
(II) What was his percentage yield?
(III) If the 32.70 g of Zinc that he started with contained impurities (materials other than Zinc), will this support the student’s percentage yield?

Explain briefly.

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Bơ 5 years 2021-08-11T17:02:56+00:00 1 Answers 10 views 0

Answers ( )

  1. Giakhanh
    0
    2021-08-11T17:04:53+00:00
    Reply

    Answer:

    (I) m_{ZnCl_{2}} = 34.07 g  

    (II) \% y = 176.4 \%

    (III) The presence of impurities in the 32.70 g of zinc will support the student’s percentage yield.

    Explanation:

    The reaction is:

    Zn + 2HCl → ZnCl₂ + H₂

    (I) To find the number of moles of ZnCl₂ we need to calculate the number of moles of the reactants.

    n_{Zn} = \frac{m}{A}

    Where:

    m: is the mass

    A: is the standard atomic weight  

    n_{Zn} = \frac{32.70 g}{65.38 g/mol} = 0.50 moles

    n_{HCl} = \frac{18.23 g}{36.458 g/mol} = 0.50 moles

    If 1 mol of Zn reacts with 2 moles of HCl, the number of moles of Zn needed to react with 0.50 moles of HCl is:              

    n_{Zn} = \frac{1 mol Zn}{2 moles HCl}*0.50 moles HCl = 0.25 moles  

    Then, the limiting reactant is HCl. Now, we can find the number of moles of ZnCl₂ by knowing that 2 moles of HCl produce 1 mol of ZnCl₂.

    n_{ZnCl_{2}} = \frac{1 mol ZnCl_{2}}{2 moles HCl}*0.50 moles = 0.25 moles    

    Finally, the mass of ZnCl₂ that he has produced is:            

    m_{ZnCl_{2}} = n_{ZnCl_{2}}*M = 0.25 moles*136.286 g/mol = 34.07 g  

    (II) The percentage yield can be found with the following equation:

    \% y = \frac{m_{e}}{m_{t}} \times 100

    Where:      

    m_{e}: is the experimental mass = 60.1 g

    m_{t}: is the theoretical mass = 34.07 g

    Hence, the percentage yield is:  

    \% y = \frac{60.1 g}{34.07 g} \times 100 = 176.4 \%

    (III) The presence of impurities in zinc will explain why the percentage yield is greater than 100% because the mass of the impurities will make the mass of zinc bigger than its mass without impurities. If there were no impurities in the zinc, its mass would be smaller and thus the percent yield.

    I hope it helps you!                  

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