Respuesta :
Answer:
percentage yield of water= [tex]\frac{1.31}{4.71} \times 100=27.8[/tex]
Explanation:
First write the balanced chemical reaction:
[tex]2CH_3(CH_2)_2CH_3(g)+13O_2(g)=8CO_2(g)+10H_2O(l)[/tex]
mass of buane=4.65g
mass of oxygen=10.8g
mass of water produced=1.31g
mole of butane=[tex]\frac{4.65}{58}=0.08[/tex]
mole of oxygen=[tex]\frac{10.8}{32}=0.34[/tex]
2 moles of butane reacts with 13 moles of oxygen;
so 1 mole of butane reacts with 6.5 moles of oxygen;
0.08 mole will react 0.52 mole of oxygen but we have only 0.34 mole;
therefore butane is excess reagent and oxygen will be the limiting reagent.
amount of product depend on the amount of limiting reagent.
so 13 mole oxygen produce 10 mole of water
hence 1 mole will produce 10/13 mole of water
0.34 mole will give [tex]\frac{10}{13} \times 0.34[/tex] mole of water.
theoretical mass of water =[tex]\frac{10}{13} \times 0.34 \times 18 =4.71 g[/tex]
but actual produced mass of water =1.31 g
[tex]percentage yield =\frac{actual yield}{theoretical yield} \times 100[/tex]
percentage yield of water=[tex]\frac{1.31}{4.71} \times 100=27.8[/tex]
