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

Answer:

2.29 L

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:

$5.98~g~Zn\frac{1~mol~Zn}{65.38~g~Zn}=0.0915~mol~Zn$

Second step:

$0.0915~mol~Zn\frac{1~mol~H_2}{1~mol~Zn}=0.0915~mol~H_2$

Third step:

We have to remember that R = 0.082 $\frac{atm*L}{mol*K}$ , so:

$V=\frac{0.082\frac{atm*L}{mol*K}*0.0915~mol~H_2*298K}{0.978~atm}$

$V=2.29~L$

I hope it helps!