Answer: 130.2 g
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Cl_2=\frac{78.9g}{71g/mol}=1.11moles[/tex]
[tex]2GaCl_3+3F_2\rightarrow 2GaF_3+3Cl_2[/tex]
According to stoichiometry :
3 moles of [tex]Cl_2[/tex] require = 2 moles of [tex]GaCl_3[/tex]
Thus 1.11 moles of [tex]Cl_2[/tex] will require=[tex]\frac{2}{3}\times 1.11=0.74moles[/tex] of [tex]GaCl_3[/tex]
Mass of [tex]GaCl_3=moles\times {\text {Molar mass}}=0.74moles\times 176g/mol=130.2g[/tex]
Thus 130.2 g of gallium chloride would react with excess fluorine to produce 78.9 g of chlorine
