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

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} Cl_2=(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=
`(2)/(3)* 1.11=0.74moles` of
`GaCl_3`

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

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