Note that NaOH has a molar mass of 40 g/mol, and Al has a molar mass of 27 g/mol. If you're not sure how to get the molar mass, you'll need a periodic table, and add the individual masses of each element present in the chemical formula.
To solve this, first we need to get the moles of each reactant. To get them, we just divide the given mass by the molar mass:
NaOH: 84.1 g / 40 g/mol = 2.1025 mol NaOH
Al: 51 g / 27 g/mol = 1.8889 mol Al
Then from the balanced reaction, we make a mole ratio of reactants:
2 mol NaOH / 2 mol Al or
2 mol Al / 2 mol NaOH
From the moles of reactant we calculated above, we get the corresponding moles of the other reactant using the mole ratio of reactants:
2.1025 mol NaOH * (2 mol Al / 2 mol NaOH) = 2.1025 mol Al
1.8889 mol Al * (2 mol NaOH / 2 mol Al) = 1.8889 mol NaOH
From here we can see that the limiting reactant is Al, because in order to completely react the 2.1025 mol NaOH, it needs 2.1025 mol Al, but in the given we only have 1.8889 mol Al.
To get the mass of excess reactant (NaOH), we subtract the moles from above,
2.1025 - 1.8889 = 0.2136 mol NaOH
mass,NaOH = 0.2136 mol * 40 g/mol = 8.544 g
To get the mass of water, same thing we'll do. We use the moles of the limiting reactant and mole ratio of reactant to get the moles of water:
1.8889 mol Al * (2 mol H2O / 2 mol Al) = 1.8889 mol H2O
Since water is 18 g/mol, just multiply the this with the moles to get the mass.
hope this helps~ `u`
If 84.1 g of sodium hydroxide are reacted with 51.0 g of aluminum in the following equation:
2NaOH + 2Al + 2H2O --> 2NaAlO2 + 3H2
a) what mass of excess reactant remains?
b) what mass of water would react?
2 answers
Thank you!!