Answer:
[tex]m_{Ga_2S_3}=2415.31gGa_2S_3[/tex]
Explanation:
Hello,
In this case, for the given chemical reaction, we can notice there is a 4:2 molar ratio between the burned moles of gallium and the yielded moles of gallium sulfide, therefore, we compute them as shown below:
[tex]n_{Ga_2S_3}=20.5molGa*\frac{2molGa_2S_3}{4molGa}=10.25mol Ga_2S_3[/tex]
Then, by using the molar mass of gallium sulfide (235.64 g/mol), we directly compute the grams:
[tex]m_{Ga_2S_3}=10.25mol Ga_2S_3*\frac{235.64gGa_2S_3}{1molGa_2S_3} \\\\m_{Ga_2S_3}=2415.31gGa_2S_3[/tex]
Best regards.
