Consider this reaction. KNO3(s)+Al(s)----->KNO2(s)+Al2O3(s) A.If 10.0 grams of p
ID: 852918 • Letter: C
Question
Consider this reaction. KNO3(s)+Al(s)----->KNO2(s)+Al2O3(s)A.If 10.0 grams of potassium nitrate is reacted with 10.0 grams of aluminum metal, what is the yeild of aluminum oxide? Please show work.
B.if 0.99 grams of aluminum oxide are actually obtained, what is the percent yeild? Please show work.
C. How many grams of the excess reactant remain once the reaction is completed? Please show work.
Consider this reaction. KNO3(s)+Al(s)----->KNO2(s)+Al2O3(s)
A.If 10.0 grams of potassium nitrate is reacted with 10.0 grams of aluminum metal, what is the yeild of aluminum oxide? Please show work.
B.if 0.99 grams of aluminum oxide are actually obtained, what is the percent yeild? Please show work.
C. How many grams of the excess reactant remain once the reaction is completed? Please show work.
KNO3(s)+Al(s)----->KNO2(s)+Al2O3(s)
A.If 10.0 grams of potassium nitrate is reacted with 10.0 grams of aluminum metal, what is the yeild of aluminum oxide? Please show work.
B.if 0.99 grams of aluminum oxide are actually obtained, what is the percent yeild? Please show work.
C. How many grams of the excess reactant remain once the reaction is completed? Please show work.
Explanation / Answer
molar mass of KNO3 = 101.1g ,molar mass of Al = 27g
moles of KNO3 = 10/101.1 = 0.09 moles
moles of Al = 10/27 = 0.37 moles
3KNO3(s)+2Al(s)----->3KNO2(s)+Al2O3(s)
From the reaction,
3 moles of KNO3 --------- 2 moles of Al
0.09 moles of KNO3 ------ (0.09*2)/3 = 0.06 moles of Al
so, here the limiting reagent is KNO3, as it is totally used up !
amount of Al formed = 0.06 * 27 =1.62g (theoritical yield)
actual yield = 0.99 g
% yield = (actual yield/ theoritical yield) * 100= (0.99/1.62) * 100 =61 %
Here the excess reactant is Al
excess moles = 0.37 - 0.06 = 0.31moles
wt = 0.31 * 27 = 8.37g