Question

Excess ammonia (NH3) is added to 5.137 grams of seaborgium hexachloride. The only products are 4.146 grams of a solid containing only seaborgium, nitrogen, and chlorine and 1.174 grams of a gas. The gas is 97.23% chlorine, and the remainder is hydrogen. The empirical formula of the gas is HCl. What fraction of the chlorine from the original compound is in the solid after the reaction?

Answer 1

Answer:

$w_{Cl}=0.5028$

Explanation:

Hello.

In this case, a partial chemical reaction can be written as:

$NH_3+SgCl_6\rightarrow HCl+solid$

Thus, the mass of chlorine in the initial seaborgium hexachloride (molar mass: 475.718 g/mol) is:

$m_{Cl}=5.137gSgCl_6*\frac{6*35.45gCl}{475.718 gSgCl_6} =2.297gCl$

Which is also the total chlorine. Moreover, the chlorine from the HCl is:

$m_{Cl\ in \ HCl}=1.174gHCl*0.9723=1.142g$

It means that the chlorine in the solid is:

$m_{Cl\ in\ solid}=2.297gCl-1.142gCl=1.155gCl$

Therefore, the required fraction (w) is computed by dividing the mass of chlorine in the solid by the mass of chlorine in the initial seaborgium hexachloride as the only source of chlorine at the beginning:

$w_{Cl}=\frac{1.155g}{2.297g}\\ \\w_{Cl}=0.5028$

Best regards.