Question

How many moles of sulfuric acid are needed to produce 57.8 milliliters of water? Water has the density of 0.987 g/mL. Show all steps of your calculation as well as the final answer.

NaOH + H2SO4 → Na2SO4 + H2O

Answer 1

mass water = 57.8 mL x 0.987 g/mL=57.0 g
moles water = 57.0 g/ 18.02 g/mol=3.17
the ratio between H2SO4 and H2O is 1 : 2
moles H2SO4 required = 3.17/2=1.58

moles MgO = 102.3 g/40.304 g/mol = 2.54
the ratio between MgO and Mg is 1 : 1
moles Mg = 2.54
mass Mg = 2.54 mol x 24.305 g/mol=61.7 g
V = mass /d = 61.7 g / 1.74 g/mL=35.5 mL

mass Fe = 78.4 mL x 7.87 g/mL=617 g
moles Fe = 617 g/ 55.847 g/mol=11.0
the ratio between Fe and FeCl3 is 1 : 1
moles FeCl3 = 11.0
mass FeCl3 = 11.0 mol x 162.206 g/mol=1784 g ☻