What mass of phosphoric acid (H_3PO_4, 98.00 g/mol) is produced from the reactio

Question

What mass of phosphoric acid (H_3PO_4, 98.00 g/mol) is produced from the reaction of 10.00 g of P_4O_10 (283.89 g/mol) with 12.00 g water (18.02 g/mol)? 10.90 g 13.81 g 43.52 g 0.14 g 14.52 g Al_2Cl_6 (a dimer of AlCl_3, 266.7 g/mol) is made by treating scrap aluminum with chlorine gas. If a reaction is run with 270 g of aluminum and 710 grams of chlorine, the limiting reactant is_____, and the theoretical yield is_____grams of Al_2Cl_6. 2 Al(s) + 3 Cl_2(g) rightarrow Al_2Cl_6(s). Al, 890 g Cl_2, 890 g Al, 1335 g Cl_2, 1335 g Cl_2, 2670 g A mass of 11.60 g of phosphoric acid (98.00 g/mol) was produced from the reaction of 10.00 g of P_4O_10 (283.9 g/mol) with 12.0 g water (18.02 g/mol). What was the percent yield for this reaction? 78.62% 101.1% 92.70% 84.00% 52.70% Write the balanced reaction equation for the decomposition of solid ammonium nitrate into nitrogen gas, oxygen gas, and water vapor. Gold metal will dissolve in an aqueous sodium cyanide solution in the presence of O_2. Balance the equation showing this process._____Au(s) +_____NaCN(aq) +_____O_2(g) +_____H_2O(l) rightarrow_____NaAu(CN)_2(aq)+_____NaOH(aq) Solid nickel(II) phosphate and aqueous sodium chloride are produced when aqueous solutions of nickel chloride and sodium phosphate are mixed. Write the balanced chemical equation. Include states of matter. Copper metal reacts with nitric acid to form aqueous copper(II) nitrate, nitrogen monoxide gas, and water. Write the balanced chemical equation. Include states of matter. Write the balanced reaction equation for the combustion of isooctane (C_8H_18), which is found in gasoline.

Explanation / Answer

Q.48: moles of P4O10 taken = mass/molar mass = 10.00 g / 283.89 g/mol = 0.035225 mol

moles of H2O taken = mass/molar mass = 12.00 g / 18.02 g/mol = 0.6660 mol

The balanced chemical reaction is

P4O10 + 6H2O ------ > 4H3PO4

From the above balanced chemcical reaction it is clear that

1 mol P4O10 reacts with 6 mol H2O

Hence 0.035225 mol P4O10 that will react with the moles of H2O = 0.035225 mol P4O10 x (6 mol H2O / 1 mol P4O10)

= 0.21135 mol H2O

However we have initially taken  0.6660 mol H2O.

Hence H2O is the excess reactant and P4O10 is the limiting reactant and P4O10 will decide the amount of H3PO4 formed.

Hence amount of P4O10 produced = 0.035225 mol P4O10 x (4 mol H3PO4 / 1 mol P4O10)

= 0.1409 mol H3PO4

=> mass of H3PO4 formed = 0.1409 mol x 98.00 g/mol = 13.81 g (answer)

Hence option (b) is correct

Q.49:

moles of Al(s) taken = mass/molar mass = 270 g / 26.98 g/mol = 10.007 mol Al

moles of Cl2(g) taken = mass/molar mass = 710 g / 70.91 g/mol = 10.0127 mol Cl2

From the balanced chemcical reaction it is clear that

2 mol Al reacts with 3 mol Cl2

Hence 10.007 mol Al that will react with the moles of Cl2 = 10.007 mol Al x (3 mol Cl2 / 2 mol Al)

= 15.011 mol Cl2

However we have initially taken only 10.0127 mol Cl2.

Hence Cl2 is completely exhausted and act as the limiting reactant.(answer)

Cl2 will decide the amount of product formed.

Hence amount of Al2Cl6 produced = 10.0127 mol Cl2. x (1 mol Al2Cl6 / 3 mol Cl2)

= 3.3376 mol Al2Cl6

=> mass of Al2Cl6 formed = 3.3376 mol x 266.7 g/mol = 890 g Al2Cl6 (answer)

Hence option (b) is correct

Q.50:

From Q.48 we calculated the theoritical yield of H3PO4 = 13.81 g

Given the actual yield of H3PO4 = 11.60 g

Hence % yield = (actual yield / theoritical yield) x 100 = (11.60 g / 13.81 g) x 100 = 84.00 %

Hence option (d) is correct.

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