estion Completion Status QUESTION 18 If 3.6 moles of Al and 3,.1 moles of Fez03
ID: 1027767 • Letter: E
Question
estion Completion Status QUESTION 18 If 3.6 moles of Al and 3,.1 moles of Fez03 are allowed to react, what is the maximum number of moles of Al203 that can be produced? 2Al Fe203-Al203+2Fe QUESTION 19 When 15.5 g of aluminum (26.98 gmol) reacts with 14.7 g of oxygen gas (32.00 g/mol), what is the maximum mass (in grams) of aluminum oxide (101.96 g/mol) that can be QUESTION 20 What is the maximum number of grams of NO (30.01 g/mol) that can be formed from the reaciton of 15.9 g of NH3 (17.03 g/mo) with 25.9 g of 02 (32.00 gmol? QUESTION 21 1.00000p t 143 meles of Cu n 40 7 mo ofINOs are alowesd to reat, how many imoles of excess reactant will remain if the reaction goes to compfetion?Explanation / Answer
Q.NO.18
2 Al + Fe2O3 ---------- Al2O3 + 2 Fe
number of moles of Al= 3.6 moles
number of moles of Fe2O3 = 3.1 moles
according to equation
1 mole of Fe2O3 = 2 mole of Al
3.1 mole of Fe2O3 = ?
= 2x3.1/1= 6.2 moles of Al
we need 6.2 mole of Al.but we have 3.6 mole of Al.so Al is completed in the reaction first.
Hence Al is limiting reagent.
according to eqiation
2 mole of Al = 1 mole of Al2O3
3.6 moles of Al= ?
= 1x3.6/2 = 1.8 mole of Al2O3
maximum number of moles of Al2O3= 1.8 moles
Q.No.19.
4 Al(s) + 3 O2(g) ------ 2 Al2O3(s)
mass of Al= 15.5 grams
molar massof Al= 26.98 gram/mole
number of moles of Al= 15.5/26.98 = 0.574 moles
mass of O2= 14.7 grams
molar mass of O2= 32.00 grams
number of moles of O2= 14.7/32.00 = 0.459 moles
according to equation
3 moles of O2= 4 moles of Al
0.459 moles of O2= ?
= 4x0.459/3= 0.612 moles of Al
we need 0.612 moles of Al.but we have 0.574 moles of Al.
so Al is limiting reagent.
according to equation
4 mole of Al= 2 moles of Al2O3
0.574 moles of Al = ?
= 2x0.574/4 = 0.287 moles of Al2O3
number of moles of Al2O3 formed= 0.287 moles
molar mass of Al2O3= 101.96 gram/mole
mass of 0.287 moles of AL2O3 = 0.287x101.96 = 29.26 grams.
maximum mass of Al2O3 formed= 29.26 grams.
Q.NO.20.
4 NH3(g) + 5 O2(g) ---- 4 NO(g) + 6 H2O(l)
mass of NH3= 15.9 grams
molar mass of NH3= 17.03 gram/mole
number of moles of NH3= 15.9/17.03= 0.934 moles
mass of O2 = 25.9 grams
molar mass of O2= 32.00 gram/mole
number of moles of O2= 25.9/32.00= 0.809 moles
according to equation
4 moles of NH3 = 5 moles of O2
0.934 moles of NH3=?
= 5x0.934/4= 1.1675 moles of O2
we need 1.1675 moles of O2. but we have 0.809 moles of O2
Hence O2 is limiting reagent
according toequation
5 mole of O2 = 4 moles of NO
0.809 moles of O2 = ?
= 4x0.809/5= 0.6472 moles
number of moles of NO formed= 0.6472 moles
molar mass of NO= 30.01 gram/mole
mass of 0.6472 moles of NO= 0.6472x30.01= 19.42 grams
mass of NO = 19.42 grams.
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