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?

Answer 1

I have the same homework and the answer is 4.80L

Answer 2

The chemical reaction is given as:

`Zn(s)+2HCl(aq)\rightarrow ZnCl_(2)(aq)+H_(2)(g)`

First, calculate the number of moles of zinc:

Number of moles =
`(Given mass in g)/(Molar mass)`

Given mass of zinc =
`14.0 g` and molar mass of zinc =
`65.4 g/mol`

Number of moles =
`(14.0 g)/(65.4 g/mol)`

=
`0.2140 moles`

Now, moles of hydrogen =
`number of moles of zinc* (1 mol of hydrogen)/(1 mol of zinc)` ( as 1 mole of zinc gives 1 mole of hydrogen)

=
`0.2140* (1 mol of hydrogen)/(1 mol of zinc)`

=
`0.2140 mol` of hydrogen.

Volume of hydrogen is calculated by:

`PV=nRT`

where, P = pressure = 1 atm at STP

V = volume

n= number of moles

R = gas constant =
`0.082 Latm/Kmol`

T = temperature= 273 K at STP

Now, insert the values in formula, we get

`V=(nRT)/(P)`

`V=(0.2140 mol* 0.082 Latm/Kmol* 273 K)/(1 atm)`

`V=(0.2140 mol* 0.082 Latm/Kmol* 273 K)/(1 atm)`

`V=4.790 L\simeq 4.80 L`

Thus, volume of hydrogen is
`4.80 L` i.e. second option is the correct answer.