Question

For the following reaction, 111 grams of perchloric acid (HCIO) are allowed to react with 30.2 grams of tetraphosphorus decaoxide. perchloric acid (HCIO 1)(aq) + tetraphosphorus decaoxide(8) phosphoric acid(aq) + dichlorine heptaoxide(1) What is the maximum amount of phosphoric acid that can be formed? grams What is the FORMULA for the limiting reagent What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group 8 more group attempts remaining

Answer 1

12HClO4 + P4O10 ---> 4H3PO4 + 6Cl2O7
mass of HCLO4 = 111 g
molar mass of HCLO4= [1*1.01+1*36.45 + 4*16]g/mol= 100.46 g/mol
number of mols of HClO4 present = 111g/100.46g/mol = 1.104P4H10 mols
mass of P4O10 =30.2 g
molar mass of P4H10 = 4*30.97 g/mol + 10*1.01 g/mol = 133.97 g/mol
number of mols of P4H10 present = 30.2 g/133.97 g/mol = 0.225 mols
mole ratio of HCLO4 : P4h10 = 12:1

So 0.225 mols P4H10 need [12*0.225 mol] HCLO4 = 2.705 mols HClO4 needed

but mols of HCLO4 present = 1.1049 mols
SInce there is fewer mols for HCLO4
Limiting reagent = HCLO4
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Thus moles of product formed depends on mols of limiting reactant

12 mol HCLO4 yields 2 mols H3PO4

So 1.1049 mol HCLO4 yields[4/12*1.10 mol] H3PO4 = 0.3683 mols
we have molar mass of H3PO4 = 98 g/mol
Mass of H3PO4 produced = number of mols of H3PO4*molar mass
= 0.3683 mol*98 g/mol = 36.09 g
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Since HCLO4 is limiting reactant , P4H10 is excess reactant
1.1049 mols HCLO4 react with [1/12*1.1049 mol of P4H10
= 0.090752 mols
Therfore leftover number of mols of P4H10 = 0.225 mols-0.092075 mol = 0.133 mols
Mass of P4H10 remaining = 0.133 mols*133.97 g/mol = 17.82 g
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