The balanced equation for the reaction is as follows
Cu₂O + 2HCl ---> 2CuCl + H₂O
Molar ratio of Cu₂O to CuCl is 1:2
mass of Cu₂O reacted - 73.5 g
Number of moles of Cu₂O reacted - 73.5 g / 143 g/mol = 0.51 mol
According to the molar ratio,
when 1 mol of Cu₂O reacts then 2 mol of CuCl is formed
therefore when 0.51 mol of Cu₂O reacts then - 2 x 0.51 mol of CuCl is formed
number of CuCl moles formed - 1.02 mol
mass of CuCl formed - 1.02 mol x 99 g/mol = 101 g
mass of CuCl formed is 101 g
The atomic mass on the periodic table represents the sum of number of protons and number of neutrons.
Atomic mass = Number of protons + number of neutrons
Hope this helps!
Ca(OH)₂ ==> Ca²⁺ + 2 OH<span>-
Ca(OH)</span>₂ is <span>strong Bases</span><span>
</span>Therefore, the [OH-] equals 5 x 10⁻⁴ M. For every Ca(OH)₂ you produce 2 OH⁻<span>.
</span>
pOH = - log[ OH⁻]
pOH = - log [ <span>5 x 10⁻⁴ ]
pOH = 3.30
pH + pOH = 14
pH + 3.30 = 14
pH = 14 - 3.30
pH = 10.7
hope this helps!</span>
<span> Ksp = [Ag+]^2[CO32-]that should be it </span>
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
