Question

What mass of Gallium chloride would react with excess fluorine to produce 78.9 g of chlorine? Gallium fluoride is the other product. 2 GaCl3 + 3 F2 2 GaF3 + 3 Cl2

Answer 1

Answer: 130.2 g

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} Cl_2=\frac{78.9g}{71g/mol}=1.11moles$

$2GaCl_3+3F_2\rightarrow 2GaF_3+3Cl_2$

According to stoichiometry :

3 moles of $Cl_2$ require  = 2 moles of $GaCl_3$

Thus 1.11 moles of $Cl_2$ will require=$\frac{2}{3}\times 1.11=0.74moles$  of $GaCl_3$

Mass of $GaCl_3=moles\times {\text {Molar mass}}=0.74moles\times 176g/mol=130.2g$

Thus 130.2 g of gallium chloride would react with excess fluorine to produce 78.9 g of chlorine