This question has multiple parts. Work all the parts to get the most points. Amm

Question

This question has multiple parts. Work all the parts to get the most points. Ammonia gas is synthesized by combining hydrogen and nitrogen: 3 H2(8) +N2(g) 2 NH3 (g) If you want to produce 696 g of NHs, what volume of H2 gas, at 53 oC and 745 mm Hg, is required? Volume- Nitrogen for this reaction will be obtained from air. of air measured at 22 °C and 745 mm Hg pressure, will be required to provide the nitrogen needed to produce 696 g of NH3? Assume the sample of air contains 78.1 mole % N2. b Nitrogen for this reaction will be obtained from air What volume of air, Volume -

Explanation / Answer

Solution :-

3H2 + N2 --- > 2NH3

Part a)

696 g NH3 need to be produced

Volume of H2 needed at 53 oC and 745 mmHg

Lets first calculate moles of H2 needed to produce 696 g NH3

(696 g NH3 * 1 mol NH3 / 17.03 g NH3)*(3 mol H2 / 2 mol NH3) = 61.3 mol H2

Now using the ideal gas law formula we can find the volume of H2

T= 53 oC +273 = 326 K

P = 745 mmHg * 1 atm / 760 mmHg = 0.980 atm

PV= nRT

V= nRT/P

V=61.3 mol * 0.08206 L atm per mol K * 326 K / 0.980 atm

V= 1673 L H2

Therefore volume of H2 needed is 1673 L we can write it as 1670 L using 3 sig fig.

Part B)

Lets calculate the moles of 696 g NH3

Moles = mass / molar mass

696 g NH3 / 17.03 g per mol = 40.9 mol NH3

Using the moles of NH3 we can find the moles of N2 needed

40.9 mol NH3 * 1 mol N2 / 2 mol NH3 = 20.45 mol N2

Using the percent of mole in the air we can find the moles of Air needed

20.45 mol N2 * 100 % / 78.1 % = 26.2 mol air

T= 22 oC +273= 295 K

P = 745 mmHg * 1 atm / 760 mmHg = 0.980 atm

Lets calculate volume of air

PV= nRT

V= nRT/ P

V= 26.2 mol air *0.08206 L atm per mol K * 295 K / 0.980 atm

V= 647 L air

Therefore the volume of air needed is 647 L

Q2) Balanced equation

Si(s)+ 2CH3Cl(g) ----- > (CH3)2SiCl2(g)

Lets calculate the moles of CH3Cl using the ideal gas law formula at 25 oC and 657 mmHg

P= 657 mmHg *1 atm / 760 mmHg = 0.8645 atm

T= 25 oC+ 273=298 K

V= 7.80 L

PV= nRT

PV/ RT= n

0.8645 atm *7.80 L /0.08206 L atm per mol K * 298 K = n

0.276 mol CH3Cl

Now lets calculate moles of Si

Moles of Si = mass / molar mass

                    = 1.98 g / 28.0855 g per mol

                    = 0.0705 molSi

Now lets calculate the moles of CH3Cl needed to react with 0.0705 mol Si using the mole ratio of the balanced equation

0.0705 mol Si * 2 mol CH3Cl / 1 mol Si = 0.141 mol CH3Cl

Moles of CH3Cl present are more than needed for the reaction therefore Si is limiting reactant hence we can find the mass of product using the limiting reactant Si

(0.0705 mol Si * 1 mol (CH3)2SiCl2 / 1 mol Si) *(129.06 g / 1 mol (CH3)2SiCl2 = 9.10 g (CH3)2SiCl2

Hence the mass of product that can be formed is 9.10 g (CH3)2SiCl2

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