In: Chemistry
Aqueous hydrochloric acid HCl will react with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid water H2O . Suppose 0.36 g of hydrochloric acid is mixed with 0.164 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.
Molar mass of HCl,
MM = 1*MM(H) + 1*MM(Cl)
= 1*1.008 + 1*35.45
= 36.458 g/mol
mass(HCl)= 0.36 g
use:
number of mol of HCl,
n = mass of HCl/molar mass of HCl
=(0.36 g)/(36.458 g/mol)
= 9.874*10^-3 mol
Molar mass of NaOH,
MM = 1*MM(Na) + 1*MM(O) + 1*MM(H)
= 1*22.99 + 1*16.0 + 1*1.008
= 39.998 g/mol
mass(NaOH)= 0.164 g
use:
number of mol of NaOH,
n = mass of NaOH/molar mass of NaOH
=(0.164 g)/(39.998 g/mol)
= 4.1*10^-3 mol
Balanced chemical equation is:
HCl + NaOH ---> H2O + NaCl
1 mol of HCl reacts with 1 mol of NaOH
for 9.874*10^-3 mol of HCl, 9.874*10^-3 mol of NaOH is
required
But we have 4.1*10^-3 mol of NaOH
so, NaOH is limiting reagent
we will use NaOH in further calculation
Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol
According to balanced equation
mol of H2O formed = (1/1)* moles of NaOH
= (1/1)*4.1*10^-3
= 4.1*10^-3 mol
use:
mass of H2O = number of mol * molar mass
= 4.1*10^-3*18.02
= 7.4*10^-2 g
Answer: 0.074 g