Respuesta :
Answer:
a)The percentage yield of the reaction is 87.3%.
b) 0.343 gram is the mass of CO that passed through.
Explanation:
[tex]I_2O_5+ 5CO\rightarrow I_2 + 5CO_2[/tex]
Moles of carbon monoxide gas = [tex]\frac{2.00 g}{28 g/mol}=0.07143 mol[/tex]
According to reaction, 5 moles of carbon monoxide gives 1 moles of iodine gas,then 0.07143 moles of carbon monoxide will give :
[tex]\frac{1}{5}\times 0.07143 mol=0.014286 mol[/tex] of iodine gas
Mass of 0.014286 moles of iodine gas:
0.014286 mol × 254 g /mol = 3.63 g
Theoretical yield of the iodine gas = 3.63 g
Experimental yield of the iodine gas = 3.17 g
Percentage yield of the reaction :
[tex]\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
[tex]=\frac{3.17 g}{3.63 g}\times 100=87.3\%[/tex]
The percentage yield of the reaction is 87.3%.
b)
Mass of iodine gas produced = 3.17 g
Moles of iodine gas = [tex]\frac{3.17 g}{254 g/mol}=0.01248 mol[/tex]
According to reaction , 1 mole of iodine gas is obtained from 5 moles of carbon monoxide gas, then 0.01248 moles of iodine gas will be obtained from :
[tex]\frac{5}{1}\times 0.01248 mol=0.06240 mol[/tex] of carbon monoxide
Moles of carbon monoxide reacted = 0.06240 mol
Moles of carbon monoxide gas used = 0.07143 mol
Moles of carbon monoxide gas which do not reacted:
0.07143 mol - 0.06240 mol = 0.00903 mol
Mass of 0.00903 moles of carbon monoxide gas:
0.00903 mol × 28 g/mol = 0.343 g
0.343 gram is the mass of CO that passed through.
