When 14.0 g of zinc metal reacts with excess hcl, how many liters of h2 gas are produced at stp? when 14.0 g of zinc metal reacts with exces

Question

When 14.0 g of zinc metal reacts with excess hcl, how many liters of h2 gas are produced at stp? when 14.0 g of zinc metal reacts with excess hcl, how many liters of h2 gas are produced at stp? 9.60 l 4.80 l 0.208 l 0.416 l?

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2022-04-10T06:58:26+00:00 0 Answers 0

Answers ( 2 )

  1. Eliza
    0
    2022-04-10T06:59:52+00:00

    I have the same homework and the answer is 4.80L

  2. Eva
    0
    2022-04-10T06:59:54+00:00

    The chemical reaction is given as:

    [tex]Zn(s)+2HCl(aq)rightarrow ZnCl_{2}(aq)+H_{2}(g)[/tex]

    First, calculate the number of moles of zinc:

    Number of moles = [tex]frac{Given mass in g}{Molar mass}[/tex]

    Given mass of zinc = [tex]14.0 g[/tex] and molar mass of zinc = [tex]65.4 g/mol[/tex]

    Number of moles = [tex]frac{14.0 g}{65.4 g/mol}[/tex]

    = [tex]0.2140 moles[/tex]

    Now, moles of hydrogen = [tex]number of moles of zinctimes frac{1 mol of hydrogen}{1 mol of zinc}[/tex] ( as 1 mole of zinc gives 1 mole of hydrogen)

    = [tex]0.2140times frac{1 mol of hydrogen}{1 mol of zinc}[/tex]

    = [tex]0.2140 mol[/tex] of hydrogen.

    Volume of hydrogen is calculated by:

    [tex]PV=nRT[/tex]

    where, P = pressure = 1 atm at STP

    V = volume

    n= number of moles

    R = gas constant = [tex]0.082 Latm/Kmol[/tex]

    T = temperature= 273 K at STP

    Now, insert the values in formula, we get

    [tex]V=frac{nRT}{P}[/tex]

    [tex]V=frac{0.2140 moltimes 0.082 Latm/Kmoltimes 273 K}{1 atm}[/tex]

    [tex]V=frac{0.2140 moltimes 0.082 Latm/Kmoltimes 273 K}{1 atm}[/tex]

    [tex]V=4.790 Lsimeq 4.80 L[/tex]

    Thus, volume of hydrogen is [tex]4.80 L[/tex] i.e. second option is the correct answer.

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