Answer: 1.98 g
Explanation:
To calculate the moles :
The balanced given equation is:
According to stoichiometry :
4 moles of 
will produce = 2 moles of 
Thus 0.22 moles of will produce=
of
Mass of 
Thus 1.98 g of water is produced along with 5.0 L of at STP
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:
The isotopic mass of 41K is 40.9574 amu
Explanation:
Step 1: Data given
The isotopes are:
39K with an isotopic mass of 38.963707u and natural abundance of 93.2581%
40K with an isotopic mass of 39.963999u
41K wit natural abundance of 6.7302 %
Average atomic mass =39.098 amu
Step 2: Calculate natural abundance of 40 K
100 % - 93.2581 % - 6.7302 %
100 % = 0.0117 %
Step 3: Calculate isotopic mass of 41K
39.098 = 38.963707 * 0.932581 + 39.963999 * 0.000117 + X * 0.067302
39.098 = 36.33681 + 0.0046758 + X * 2.067302
X = 40.9574 amu
The isotopic mass of 41K is 40.9574 amu
