Answer:
[tex]m_{HI}=860gHI\\\\Y=90.1\%[/tex]
Explanation:
Hello.
In this case, for the reaction:
[tex]H_2+I_2\rightarrow 2HI[/tex]
As 6.79 g of hydrogen (molar mass = 2.02 g/mol) react in excess iodine, we can compute the theoretical yield of hydrogen iodide (molar mass = 127.91 g/mol) via their 1:2 mole ratio shown on the chemical reaction:
[tex]m_{HI}=6.79gH_2*\frac{1molH_2}{2.02gH_2}* \frac{2molHI}{1molH_2}*\frac{127.91gHI}{1molHI}\\\\m_{HI}=860gHI[/tex]
Next, we compute the percent yield by divinding the actual yield (775 g) by the theoretical yield (860 g):
[tex]Y=\frac{775g}{860g}*100\%\\\\Y=90.1\%[/tex]
Best regards!