Question

Percent yield of chemical reactions Aqueous hydrobromic acid (HBr) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid water (HO). 189.71 g of water is produced from the reaction of 79.3 g of hydrobromic acid and 70.2 g of sodium hydroxide, calculate the percent yield of water. Round your answer to 3 significant figures. x 5 ?

Answer 1

NaOH(s) + HBr (aq) ---> NaBr + H20

given
mass of HBr =79.3 g
Mass of NaOH = 70.2 g

we have molar mass of HBr = [1*1.01 + 1*79.9] g/mol =80.91 g/mol
molar mass of NaOH = [1*22.99 +1*16+1*1.01]g/mol =40 g/mol

Now let us find out

moles of HBr present= mass of HBr/molar mass of HBr
= 79.3 g/80.91 g/mol = 0.98 mols

moles of NaOH present = mass of NaOH/Molar mass of NaOH
=70.2g/40 g/mol = 1.755 mols

According to balanced equation mole ratio NaOH:HBr = 1:1
And number of moles of HBr present is fewer than the mols of NaOH
present .Hence the reaction will continue until all HBr consumed
Thus
limiting reactant is HBr
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SInce HBr is the limiting reactant, moles of product formed depends on
mols of HBr
SInce mols of HBr : H2O = 1:1
moles of H2O produced = mols of HBr present = 0.98 mols
Molar mass = 18.02 g/mol
mass of water formed = mols of water present*molar mass
=0.98 mols *18.02 g/mol = 17.6596 g

This is the theoretical yield of water
But actual yield = 9.71 g
Percent yield = actual yield/theoretical yield *100 =9.71/17.6596 *100
= 54.98 %
rounding for 3 sig figures
Answer = 55.0 %
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