Question

zinc metal and hydrochloric acid react together according to the following equation 2HCl (aq)+ Zn(s) = ZnCl2(aq) +H2(g) if 5.98 g Zn reacts with excess HCl at 298 K and 0.978 atm what volume of H2 can be collected

Answer 1

2.29 L

Step-by-step explanation:

In this question we have to start with the chemical reaction:

`2HCl_(_a_q_)~+~Zn_(_s_)~->~ZnCl_2_(_a_q_)~+~H_2_(_g_)`

The reaction is already balanced. So, if we have an excess of HCl the compound that would limit the production of
`H_2` would be Zn. So, we have to follow a few steps:

1) Convert from grams to moles (Using the atomic mass of Zn 65.38 g/mol).

2) Convert from moles of Zn to moles of
`H_2` (Using the molar mass 1 mol
`H_2` = 1 mol Zn).

3) Convert from mol of
`H_2` to volume (Using the ideal gas equation PV=nRT).

First step:

`5.98~g~Zn(1~mol~Zn)/(65.38~g~Zn)=0.0915~mol~Zn`

Second step:

`0.0915~mol~Zn(1~mol~H_2)/(1~mol~Zn)=0.0915~mol~H_2`

Third step:

We have to remember that R = 0.082
`(atm*L)/(mol*K)` , so:

`V=(0.082(atm*L)/(mol*K)*0.0915~mol~H_2*298K)/(0.978~atm)`

`V=2.29~L`

I hope it helps!