The reaction between metallic Magnesium and steam will be
As per balanced equation:
one mole of Mg will react with two moles of water to give one mole of magnesium hydroxide
Moles of Mg = Mass / atomic mass = 19.4/24 = 0.81
Moles of water = mass / molar mass = 14/18 =0.78
thus here for 0.81 moles of Mg we need twice the moles of water
However we have just 0.78 moles of water it means the limiting reagent is water
The moles of Mg reacted with water = 0.5 X 0.78 = 0.39 moles
moles of Mg left after reaction = 0.81-0.39= 0.42 moles
mass of Mg left (excess reagent) = moles X atomic mass = 0.42 X 24 = 10.08g
The moles of Mg(OH)2 formed = 0.39 moles
mass of Mg(OH)2 formed = moles X molar mass = 0.39 X 58 = 22.62g
They are almost the same size and they are rocky planets (like not made of gas)
Hope this helps :)
The acid and alkali that would react to make sodium sulfate is
alkali= sodium hydroxide
acid = sulphuric acid
<u><em>explanation</em></u>
Acid react react with alkali to form salt and water.
For sodium sulfate salt to be formed a sulfate acid is required to react with a hydroxide of sodium .
therefore Sodium hydroxide (alkali) react with sulphuric acid (an acid) to form sodium sulfate (salt) and water.
The equation for the reaction between NaOH and AlCl₃ is as follows;
3NaOH + AlCl₃ ---> 3NaCl + Al(OH)₃
the stoichiometry of NaOH : AlCl₃ is 3:1
3 moles of NaOH reacts with 1 mol of AlCl₃ to produce 1 mol of Al(OH)₃
the number of AlCl₃ moles reacted - 6.5 mol
molar mass of NaOH -(23 +16 +1) = 40 g/mol
the number of NaOH moles reacted = 57.0 g / 40 g/mol
NaOH moles = 1.425 mol
either NaOH or AlCl₃ is in excess and other is the limiting reactant.
limiting reactant is the reactant whose number of moles are fully consumed during the reaction. the reactant that is in excess will have leftover moles that are remaining after the reaction.
If AlCl₃ is the limiting reactant, number of NaOH moles would be thrice the amount of AlCl₃ present,
then number of NaOH moles that should be present - 6.5 * 3 = 19.5 mol
however there are only 1.425 mol of NaOH present, therefore AlCl₃ is in excess.
Then NaOH is the limiting reactant,
the amount of products formed depends on the amount of the limiting reactant present.
stoichiometry of NaOH : Al(OH)₃ is 3:1
the number of Al(OH)₃ moles produced = number of NaOH moles reacted / 3
number of Al(OH)₃ moles are - 1.425 mol /3 = 0.475 mol
molar mass of Al(OH)₃ = (27 +3*16 + 3*1) = 78 g/mol
mass of Al(OH)₃ produced = 78 g/mol * 0.475 mol = 37.05 g