Chemical Engineering: Stoichiometry and Mass Balance

Please see the attached file for the fully formatted problems. Just do A3 and A4, please.

A3. Solid calcium fluoride (CaF2) reacts with sulphuric acid to form solid calcium sulphate (CaSO4) and gaseous hydrogen fluoride (HF). The HF is then dissolved in water to form hydrofluoric acid. A source of calcium fluoride is fluorite ore containing 96% w/w CaF2 and 4% w/w SiO2.
In a typical hydrofluoric acid manufacturing process, fluorite ore is reacted with 93% w/w aqueous sulphuric acid, supplied in 15% excess of the stoichiometric amount. Ninety five percent (95%) of the CaF2 reacts with the acid. Some of the HF formed reacts with the dissolved silica (SiO2) in the reaction. The unreacted HF gas exiting from the reactor is subsequently dissolved in enough water to produce 60% w/w aqueous hydro fluoric acid.

(a) Determine the mass of 93%w/w acid in the feed.

(b) Calculate the quantity of fluorite ore needed to produce 2500 kg of 60% w/w acid. Assume all the SiO2 present in the ore reacts with HF. The reaction equations are:

CaF2 (s) + H2SO4(aq) ---> CaSO4(s) + 2 HF(g)
6 HF(g) + SiO2(aq) ---> H2SiF6(s) + 2 H2O(l)

DATA
RAM
Ca = 40, F = 19, H = 1, S = 32, O = 16, Si = 28

A4. Formaldehyde (CH2O) is manufactured by the catalytic oxidation of methanol using an excess of air. Formic acid (HCOOH) is also formed if conditions are not properly controlled. The reaction equations are as follows:

Main reaction

CH3OH + ½ O2 → CH2O + H2O --------- (1)
Secondary reaction

CH2O + ½ O2 → HCOOH --------- (2)

The product gases in a test run have the following composition:

Component Mol %
CH3OH 8.6
CH2O 3.1
HCOOH 0.6
H2O 3.7
O2 16.0
N2 68.0

Determine the following:

(a) The molar ratio of feed air to feed methanol

(b) The per cent excess air based on the main reaction

(c) The conversion of methanol

(d) The yield of formaldehyde

(e) The selectivity of formaldehyde