35. How many moles of gas are in a gas sample occupying 1.63 L at 687 mmHg and 2

Question

35.

How many moles of gas are in a gas sample occupying 1.63 L at 687 mmHg and 293 K?

A)

16.3 mol

B)

46.6 mol

C)

0.0613 mol

D)

0.00503 mol

E)

3.82 mol

38. What volume of ammonia gas, measured at 847.0 mmHg and 58.8oC, is required to produce 5.16 g of ammonium sulfate according to the following balanced chemical equation?

2NH3(g) + H2SO4(aq) (NH4)2SO4(s)

A)

7.41 L

B)

0.000799 L

C)

1.91 L

D)

0.00320 L

E)

0.477 L

35.

How many moles of gas are in a gas sample occupying 1.63 L at 687 mmHg and 293 K?

A)

16.3 mol

B)

46.6 mol

C)

0.0613 mol

D)

0.00503 mol

E)

3.82 mol

Explanation / Answer

35 .

Given: Volume = 1.63 L , P= 687 mmHg , T = 293 K

We use ideal gas law to calculate mol

pV = nRT

p is pressure in atm, V is volume in L , n – number of mol

R is gas constant = 0.08206 L atm per K per mol

T = temperature in K

Lets convert pressure in atm

P = 687 mmHg * 1 atm/ 760 mmHg

= 0.904 atm

Number of mol = pV/ RT

=0.904 atm * 1.63 L / ( 0.0806 Latm /( K mol) * 293 K )

=0.0613 mol

So answer is C) 0.0613 mol

38. )

Given:

Pressure of NH3 = 847.0 mmHg

T = 58.8 deg C , 5.16 g (NH4)SO4

We calculate moles of ammonium sulfate

Mole of ammonium sulfate = 5.16 g / 132.1392 g per mol

= 0.0390 mol ammonim sulfate

Balanced reaction:

2NH3(g) + H2SO4(aq) ---- > (NH4)2SO4(s)

From above reaction, we say that 2 mol of NH3 required to react with 1 mol of ammonium sulfate

Number of moles of NH3 required to react with 0.0390 moles of ammonium sulfate

= 0.0390 mol ammonium sulfate * 2 mol NH3 / 1 mol of ammonium sulfate

= 0.0780 mol NH3

Now we use ideal gas law to calculate volume of NH3

Lets convert given T and pressure to K and atm

p = 847.0 mmHg * 1 atm/ 760 mmHg

= 1.11 mmHg

T in K

=58.8 deg C + 273.15 K = 331.95 K

pV = nRT

V = nRT/p

= 0.0780 mol * 0.08206 L atm ( K mol) * 331.94 K / 1.11 atm

= 1.91 L

And the option C is correct

C) 1.91 L

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