Question

A sample of gas contains 0.1500 mol of HCl(g) and 7.500×10-2 mol of Br2(g) and occupies a volume of 9.63 L. The following reaction takes place: 2HCl(g) + Br2(g)2HBr(g) + Cl2(g) Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.

Answer 1

9.63 L.

Step-by-step explanation:

Hello,

In this case, the undergoing chemical reaction is:

`2HCl(g) + Br_2(g)\rightarrow 2HBr(g) + Cl_2(g)`

So the consumed amounts of hydrochloric acid and bromine are the same to the beginning based on:

`n_(Br_2)^(consumed)=0.1500molHCl*(1molBr_2)/(2molHCl)=0.075molBr_2`

In such a way, the yielded moles of hydrobromic acid and chlorine are:

`n_(HBr)=0.1500molHCl*(2molHBr)/(2molHCl)=0.1500molHBr \\n_(Cl_2)=0.1500molHCl*(1molCl_2)/(2molHCl)=0.075molCl_2`

Thus, the volume of the sample, after the reaction is the same as no change in the total moles is evidenced, that is 9.63L.

Best regards.