Review Topics] he References to access important values if needed for this quest

Question

Review Topics] he References to access important values if needed for this question or the following reaction, 113 grams of perchloric acid (HCIO4) are allowed to react with 30.7 grams of etraphosphorus decaoxide perchloric acid (HCIO4)(aq) + tetaphosphorus decaoxide(s) phosphoric acid aq) + dichlorine heptaondel) What is the maximum amount of phosphoric acid that can be formed? grams What is the FORMULA for the limiting reagent? grams Wht amout fth ecstr 7 more group attempts remaining Retry Entire Group Submit Answer

Explanation / Answer

Tetraphosphorus decoxide formula is P4O10

Dichlorine heptoxide formula is Cl2O7

HClO4 + P4O10 = H3PO4 + Cl2O7

The balanced equation is

12 HClO4 + P4O10 = 4 H3PO4 + 6 Cl2O7

In the balanced reaction equation

12 moles of HClO4 reacts with 1 mole of P4H10 to produce 4 moles of H3PO4 and 6 moles of Cl2O7.

Now,

113 g of HClO4

Molar mass of HClO4 = 100 g/mol

So, 100 g of HClO4 = 1 mol

1 g of HClO4 = (1/100) mol

113 g of HClO4 = (113/100) mol = 1.13 mol

30.7 g of P4O10

Molar mass of P4O10 = 284 g/mol

284 g of P4O10 = 1 mol

1 g of P4O10 = (1/284) mol

30.7 g of P4O10 = (30.7/284) mol = 0.11 mol

Since, moles of HClO4 is less [ 1.13 < (12 x 0.11) ] and hence is limiting reagent.

Again,

12 HClO4 + P4O10 = 4 H3PO4 + 6 Cl2O7

In the reaction equation,

12 mole of HClO4 produces 4 moles of H3PO4

1 mole of HClO4 produces (4/12) moles of H3PO4

1.13 mole of HClO4 produces 1.13 (4/12) moles of H3PO4

1.13 mole of HClO4 produces 0.38 moles of H3PO4

Molar mass of H3PO4 = 98 g/mol

So, 1 mole of H3PO4 = 98 g

0.38 moles of H3PO4 = 0.38 x 98 g = 37.24 g

Perchloric acid is the limiting reagent with a formula is HClO4

Again

12 HClO4 + P4O10 = 4 H3PO4 + 6 Cl2O7

In the reaction equation,

12 mole of HClO4 reacts with 1 moles of P4O10

1 mole of HClO4 reacts with (1/12) moles of P4O10

1.13 mole of HClO4 reacts with (1.13/12) moles of P4O10

1.13 mole of HClO4 reacts with 0.094 moles of P4O10

Now,

Initial moles of P4O10 = 0.11 mol

moles of P4O10 reacted = 0.094 mol

So, excess moles of P4O10 left = (0.11 – 0.094) mol = 0.016 mol

Molar mass of P4O10 = 284 g/mol

1 mol of P4O10 = 284 g

0.016 mol of P4O10 = 0.016 x 284 g = 4.54 g remain

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