Hi, Can you show the step process for the following equations using the above su

Question

Hi, Can you show the step process for the following equations using the above subject method in solving each equation.. Thanks.

1.) What mass of carbon monoxide is produced from 50.0 liters of carbon dioxide? CO2 + C -> CO

2.) What volume of nitrogen gas is needed to react completely with 150.0 liters of hydrogen in the production of ammonia?

3.) What volume of oxygen gas (Liters) is produced by the decomposition of 250.0 grams of Aluminum Oxide.

4.) What volume of carbon dioxide (Liters) is produced from the reaction of 35.5 grams baking soda (NaHCO3) with hydrochloric acid? NaHCO3 + HCl -> NaCl + CO2 + H2O

5.) How many liters of ammonia (NH3) are required to produce 75 L of nitrogen monoxide (NO) gas? 4NH3 + 5O2 -> 4NO + 6H2O

6.) Oxygen and aluminum combine in a synthesis reaction. a) If 56 liters of oxygen are reacted with 40.0 grams of aluminum, what is the limiting reactant? How many grams of Aluminum oxide are produced?

Explanation / Answer

1] 22.4 L =1 mole of gas

so moles of CO2 = 50/22.4 =2.232 moles

from 1 mole of CO2 1 mole of CO is formed , so 2.232 moles of CO formed from 2.232 moles of CO2

2] N2   + 3H2 .....> 2NH3

1 mole of N2 require 3 moles of H2

   and moles proportional to volume

so 50 L will react with with 150 L   ( 1 mole require 3 mole of H2)

3] Al2O3 ----- > 2Al   +   [3/2]O2

1 mole of Al2O3 produces 1.5 moles of O2

moles of Al2O3 =mass/ MW =250/101.96 =2.4519 moles

so moles of O2 produced = 2.4519*1.5=3.678 moles

so volume of O2 = 22.4*3.678 =82.385 L

4]NaHCO3 + HCL = NaCL + H2O + CO2

1 mole of baking soda produces 1 mole of CO2

moles of NaHCO3 =35.5 /84 =0.4226 moles

so moles of CO2 produced = 0.4226

    volume of CO2 =22.4*0.4226=9.466L

5]       4NH3 + 5O2 -> 4NO + 6H2O

           1 mole NO requires 1 mole of ammonia

so to produce 1 L of ammonia required to produce 1 mole of NO

so 75 L of NH3 required to produce 75 L of NO

6]   2Al   +   [3/2]O2 ------->Al2O3

56 L O2 contains mole =56/22.4 =2.5 moles

moles of Al =40/26.9= 1.4825 mole

as 1mole of Al requires 3/4 mole O2

here here moles ofO2 higher than required so Al is limiting reagent

from 1 mole of Al   0.5 moles of Al2O3 produced

so from 1.4825 moles of Al

         Al2O3 produced = 1.4825/2 =0.74125 moles

so mass produced =mole * MW = 0.74125*101.96=75.577 gm

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.