The following two reactions are important in the blast furnace production of iro

Question

The following two reactions are important in the blast furnace production of iron metal from iron ore (Fe,0,): ce) +02(g)--200 Feao,(s) + 3CO(g) 2Fe(s) + 3co,(g) Using these balanced reactions, how many moles of O, are required for the production of 3.19 kg of Fe? Nitric oxide, NO, is made from the oxidation of NH,, and the reaction is represented by the equation 4NH +50,4NO6H2O What mass of NO can be produced from 7.55 g of NH? For the reaction P,Olds) + 6H20(/)4HPO4(ag), what mass of P,010 must be consumed if 3.71 10 molecules of H,O are also consumed? Phosphoric acid can be prepared by reaction of sulfuric acid with phosphate rock" according to the equation: Ca (PO)2 + 3H,SO,03CaSO2H,PO Suppose the reaction is carried out starting with 129 g of Ca (PO and 97.4 g of H,SO Which substance is the limiting reactant?

Explanation / Answer

ns. #1. Moles of Fe to be produced = Mass/ Molar mass

                                                = (3.190 x 103 g) / (55.847 g/mol)

                                                = 57.12035 mol

# According to stoichiometry of balanced reactions-

            I. 1 mol O2 produces 2 mol CO      (Reaction 1)

            II. 3 mol CO produces 2 mol Fe      (Reaction 2).

Comparing reaction 1 and 2-

            1 mol O2 produces 2 mol CO (reaction 1) ; and 2 mol CO would produce (2/3 = 0.6667) mol Fe (reaction 2).

So, moles of O2 required = (1 mol O2 / 0.6667 mol Fe) x Moles of Fe to be produced

                                    = (1 mol O2 / 0.6667 mol Fe) x 57.12035 mol Fe

                                    = 38.08214 mol O2

#2. Moles of NH3 = Mass / Molar mass

                                    = 7.55 g / (17.03056 g/ mol)

                                    = 0.44332 mol

According to stoichiometry of balanced reaction, 4 mol NH3 produces 4 mol NO.

So,

Moles of NO produced = 0.44332 mol = Moles of NH3 consumed

Now,

Mass of NO produced = Moles of NO produced x Molar mass

                                                = 0.44332 mol x (30.00614 g/ mol)

                                                = 13.30234 g

#3. Moles of H2O consumed = Moles of H2O consumed/ Avogadro number

                                                = (3.71 x 1023 molecules) / (6.022 x 1023 molecules/ mol)

                                                = 0.61607 mol

According to the stoichiometry of balanced reaction, 1 mol P4O10 consumed 6 mol H2O.

Now,

Moles of P4O10 consumed = (1 / 6) x Moles of H2O consumed

                                                = (1 / 6) x 0.61607 mol

                                                = 0.10268 mol

Now,

Mass of P4O10 consumed = Moles of P4O10 consumed x Molar mass

                                                = 0.10268 mol x (283.889048 g/ mol)

                                                = 29.1495 g

#4. In balanced reaction, the theoretical molar ratio of reactants is-

            Moles of Ca3(PO4)2 : moles of H2SO4 = 1 : 3

# Moles of Ca3(PO4)2 taken = 129.0 g / (310.176724 g/ mol) = 0.41589 mol

Moles of H2SO4 taken = 97.4 g / (98.07948 g/ mol) = 0.99307 mol

Now,

Experimental molar ratio of reactants =

                                    Moles of Ca3(PO4)2 : moles of H2SO4

= 0.41589 mol : 0.99307 mol

= 1 : 2.4

Comparing the theoretical and experimental molar ratio of reactants, the moles of H2SO4 are less than theoretical 3.0 mol while that of Ca3(PO4)2 is kept constant at 1.0 mol.

Hence, H2SO4 is the limiting reactant.

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