4a. Calcium cyan amide, CaCN2, (cyanamide = H2N-CEN) and calcium cyanamideCa2+ +

Question

4a. Calcium cyan amide, CaCN2, (cyanamide = H2N-CEN) and calcium cyanamideCa2+ + NEC-N2.CaCN2), is solid. It reacts with gaseous O2 under strong heating to give CaO, a solid, and two gases 2 CaCN2(s) + 7 Oza) 2 cao(s) + 2 CO2(g) + 4 NO2(g) A reaction consumes 1.40-L of oxygen (measured at 25.0°C and 1 atmosphere). Predict the volume of CO2 and the volume of NO2 present when the products are examined under the same conditions of P and T. Volume CO2 = Volume NO2_ 4b. You are asked to design an air bag for a car. You know that the bag should be filled with gas with a pressure slightly greater than atmospheric pressure and at room temperature. What quantity of sodium azide, NaN; (in both moles and grams) should be used to generate 45.5-L of nitrogen at 829-mm of Hg and a temperature of 22.0°C? The gas producing reaction is shown below. 2 NaN3(s) 2 Na(s) + 3 Nag) Moles =- Grams . 4c. A sample of methane gas having a volume of 2.80-L at 25.0°C and 1.65-atmospheres was mixed with a sample of oxygen having a volume of 35.0-L at 31.0°C and 1.25- atmospheres. The mixture was ignited to form carbon dioxide and water. Calculate the volume of CO2 formed at a pressure of 2.50-atmospheres and a temperature of 125°C CH4(g) +20mg) CO2(z) + 2 H2O(g) Limiting reagent- Moles of limiting reagents Volume CO2

Explanation / Answer

1.

Moles of oxygen (n) can be calculated from gas law equation as =PV/RT

P= pressure in atm= 1 atm, V= Volume of oxygen =1.4L and Temperature =25 deg.c= 25+273= 298K

Since Pressure and temperature remains constant, volume is proportional to no of moles

As per the reaction 2CaCN2+7O2(g)------->2CaO(s)+2CO2(g)+4NO2(g)

7 moles of oxygen gives rise to 2 moles of CO2 and 4 moles of NO2

1.4 L of oxygen hence gives 2*1.4/7 L of CO2= 0.4L and 4*1.4/7 =0.8L of NO2

2.

From gas law, PV= nRT, P = pressure in atm, 760 mm Hg= 1 atm and 829 mm Hg= 829/760 atm =1.1 atm and T= 22 deg.c= 22+273= 295K, V= 45.5 L, R= gas constant = 0.0821 L.atm/mole.K

No of moles of Nitrogen (n)= PV/RT= 1.1*45.5/(0.0821*295)= 2.1 moles

From the reaction 2NaN3(s) ----->2Na(s)+3N2(g)

2 mole of N2 requires 3 moles of NaN3

2.1 moles of N2 requires 3*2.1/2= 3.15 moles of NaN3

Molar mass of NaN3= 23+14*3= 65 g/mole, mass of NaN3= moles* molar mass= 3.15*65=204.75 gm

3.

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