Answer:
d. 12.3 grams of Al2O3
Explanation:
The balanced chemical equation of this chemical reaction is as follows:
4Al + 3O2 --> 2Al2O3
Based on the balanced equation, 4 moles of aluminum (limiting reagent) reacts to form 2 moles of aluminum oxide (Al2O3).
First, we need to convert the mass of aluminum to moles using the formula;
mole = mass/molar mass
Molar mass of Al = 27g/mol
mole = 6.50/27
= 0.241mol of Al.
Hence, if 4 moles of aluminum (limiting reagent) reacts to form 2 moles of aluminum oxide (Al2O3).
Then, 0.241mol of Al will produce (0.241 × 2/4) = 0.241/2 = 0.121mol of Al2O3.
Convert this mole value to molar mass using mole = mass/molar mass
Molar mass of Al2O3 = 27(2) + 16(3)
= 54 + 48
= 102g/mol
mass = molar mass × mole
mass = 102 × 0.121
mass of Al2O3 = 12.34grams.
1. 26
2. 115
THIS IS TO REACH 20 CHARACTERS
Answer:
8.18K
Explanation:
This question is solved using the ideal gas equation:
PV=nRT
where,
P= pressure
V= volume
n= no. of moles
R= gas constant(8.314 J/K/mol)
T= temperature
Since in the question, volume of the container and amount of gas is not given,we'll assume a value for these variables and substitute it in our formula;
P= 3.4 atm
V= 1 litre (assumption)
n= 5 moles (assumption)
R= 8.314
T= ?
PV=nRT
3.4*1=5*8.314*T
T=3.4*1/0.08314*5
T= 8.18K
