Answer:
5.5 atm
Explanation:
Step 1: Calculate the moles in 2.0 L of oxygen at STP
At STP, 1 mole of an ideal gas occupies 22.4 L.
2.0 L × 1 mol/22.4 L = 0.089 mol
Step 2: Calculate the moles in 8.0 L of nitrogen at STP
At STP, 1 mole of an ideal gas occupies 22.4 L.
8.0 L × 1 mol/22.4 L = 0.36 mol
Step 3: Calculate the total number of moles of the mixture
n = 0.089 mol + 0.36 mol = 0.45 mol
Step 4: Calculate the pressure exerted by the mixture
We will use the ideal gas equation.
P × V = n × R × T
P = n × R × T / V
P = 0.45 mol × (0.0821 atm.L/mol.K) × 298 K / 2.0 L = 5.5 atm
Answer:
163.2g
Explanation:
First let us generate a balanced equation for the reaction. This is shown below:
4Al + 3O2 —> 2Al2O3
From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.
From the equation,
4moles of Al produced 2moles of Al2O3.
Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.
Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:
Mole of Al2O3 = 1.6mole
Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol
Mass of Al2O3 =?
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass of Al2O3 = 1.6 x 102 = 163.2g
Therefore the theoretical of Al2O3 is 163.2g
Answer:
See the answer below , please.
Explanation:
In a decomposition reaction, a certain compound is "broken" to give two or more different products.
An example for compound AB, giving as products A and B:
AB -> A + B
In the case of water:
2H20 -> 2H2 + 02, water decomposes giving Hydrogen and Oxygen
N=m(g)/m.wt
n=85/12(1)+16(2) =1.93 moles
