Question

For the following reaction, 55.9 grams of diphosphorus pentoxide are allowed to react with 24.7 grams of water diphosphorus pentoxide (s) + water (1) phosphoric acid (aq) What is the maximum amount of phosphoric acid that can be formed? What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

Question 1

Balanced equation:
P2O5 + 3 H2O ====> 2 H3PO4
Reaction type: synthesis

Mass of P2O5 = 55.9 gm

Molar mass of P2O5 = 141.94 g/mol

Moles of P2O5 = 55.9 gm / 141.94 g/mol = 0.3938 Moles

Mass of water = 24.7 m

Molar mass of water = 18.02 g/mol

Moles of water = 24.7 gm / 18.02 g/mol = 1.371 Moles

P2O5 is limiting reagent here.

Moles of Phosphoric acid produced =0.7876 Moles

molar mass of Phosphoric acid = 97.994 g/mol

Mass of Phosphoric acid produced =  0.7876 mol x 97.994 g/mol =  77.184 g

Excess of water left after reaction = 24.7 - 21.284 = 3.416 gm

Balanced equation: P205 + 3 H20 = 2 H3PO4 Reaction type: synthesis Reaction stoichiometry Limiting reagent Compound Coefficient Molar Mass Moles Weight P205 141.944524 6.3938159 55.90001 HO 18.01528 1.181448 21.28411 H2PO4 97.995182 0.7876318 77.18412 Units: molar mass - g/mol, weight - g.

Question 2

Balanced equation
CO + H2O ====> CO2 + H2
Reaction type: double replacement

Mass of water = 5.73 g

Molar mass of water = 18.02 g/mol

Moles of water = 5.73 gm / 18.02 g/mol = 0.31806 Moles

Moles of CO2 produced =0.31806 Moles

Molar mass of CO2 =44.01 g/mol

Mass of CO2 produced =  0.31806 mol x 44.01 g/mol =  13.997 g

actual yield of CO2 = 9.36 g

Percentage yield of the reaction = 9.36 g x 100 % / 13.997 g = 66.87 %

Question 3

Balanced equation:
CH4 + CCl4 ====> 2 CH2Cl2
Reaction type: synthesis

Mass of carbontetra chloride = 5.73 g

Molar mass of carbontetra chloride = 153.82 g/mol

Moles of carbontetra chloride reacted = 5.73 gm / 153.82 g/mol = 0.037250 Moles

Moles of CH2Cl2 produced = 0.07450 Moles

Molar mass of CH2Cl2 = 84.93 g/mol

Mass of CH2Cl2 produced =  0.07450 mol x 84.93 g/mol =  6.327 g

Percentage yield of CH2Cl2 = 66.9 %

actual yield of the reaction = 6.327 g x 66.9 / 100 = 4.232 %

Balanced equation: CH4 + CC14 = 2 CH2Cl2 Reaction type: synthesis Reaction stoichiometry Limiting reagent Compound Coefficient Molar Mass Moles Weight CH4 16.04246 0.03725068 0.5975926 CCIA 153.8227 0.03725068 5.730000 CH2Cl2 84.93258 0.07450136 6.327593 Units: molar mass - g/mol, weight - g.