Quiz 4-26-12

Answers

2.30 g Na are reacted with 100.0 mL of 0.10 M phosphoric acid. Calculate the number of liters of hydrogen gas formed at STP.

6 Na + 2H3PO4 3 H2 + 2 Na3PO4

2.30 g Na 1 mole Na

23.0 g Na

3 mole H2

6 mole Na

22.4 L H2

1 mole H2

= 1.12 L H2

2.30 g Na are reacted with 100.0 mL of 0.10 M phosphoric acid. Calculate the number of liters of hydrogen gas formed at STP.

6 Na + 2H3PO4 3 H2 + 2 Na3PO4

0.100 L H3PO4 0.1 mole H3PO4

1L H3PO4

3 mole H2

2 mole H3PO4

22.4 L H2

1 mole H2

= 0.336 L H2

When you actually perform the experiment, you get 0.30 L of H2. What is the percent yield.

• The two values for the volume of H2 were 0.336 L and 1.12 L of H2

• 0.336 L is the theoretical yield. 1.12 is just wishful thinking.

• 0.30 L is the actual yield.

Actual yield

theoretical yieldX 100 %

0.30 L H2

0.336 L H2

= 89%

A 0.336 L of sample of hydrogen gas at STP is heated to 23oC at 745 mm Hg.Calculate the volume of the gas at room temperature.

Identify the variablesV1 = 0.336 LT1 = 273 KP1 = 760 mm HgT2 = 23 + 273 = 296 KP2 = 745 mm HgV2 = ?

P1V1

=

P2V2

T1 T2

V2

P1 V1 T2

T1P2

(760 mm Hg)

=

(0.336 L) (296 K)

(273 K)(745 mm Hg)

0.372 L =