A sample of pure sodium carbonate with a mass of 0.452 g was dissolved in water and reacted with 42.5 ml of 0.250 M hydrochloric acid. Write the balanced equation and determine the percent yield when the volume of the gas product collected at 0 oC and 760. mm Hg was 89.7 ml.

0 oC & 760. mm Hg = STP

Na2CO3 (aq) + 2 HCl (aq) → 2 NaCl (aq) + H2O(l) + CO2 (g)

0.452 g 1 mol = .00426 mol 106.0 g

0.0425 L x 0.250 M = 0.0106 mol

1 CO2 = 0.00426 mol CO2

1 Na2CO3

1 CO2 = 0.00531 mol CO2

2 HCl

@ STP : 22.4 L mol

x 0.00426 mol CO2 = 0.0954 L

0.452 g 42.5 ml 0.250 M

0 oC = 273 K 760. mm Hg = 1 Atm

89.7 ml = 93.9% 95.5 mL

R = 0.0821 L atm/mol Kor use PV = nRT

n = 0.00426 mol CO2

V = (0.00426) (0.0821) (273) = 0.0955 L = 95.5 mL1.00

Determine the percent yield when 432 mL of gas is obtained at 27oC and 700. torr by reacting 5.00 g of zinc metal with 100. mL of 0.350 M HCl.

0 oC & 760. mm Hg = STP

Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g)

5.00 g 100. ml 0.350 M

5.00 g 1 mol = .0765 mol 65.4 g

0.100 L x 0.350 M = 0.0350 mol

1 H2 = 0.0765 mol H2

1 Zn

1 H2 = 0.0175 mol H2

2 HCl

@ STP : 22.4 L mol

x 0.0175 mol H2 = 0.392 L

27+ 273 = 300. K 700./760. = 0.921 atm

432 ml = 92.3% 468 mL

or use PV = nRT

V = (0.0175) (0.0821) (300) = 0.468 L = 468 mL0.921

Using P1V1 = P2V2 : T1 T2

1 atm(0.392 L) = (0.921 atm)V2 273 K 300. K

V2 = 0.468 L

n = 0.0175 mol H2

Honors Chemistry - Equations & Constants

pH = - log[H+] pOH = - log[OH-] pH + pOH = 14

q = m ∆T c (Heat = mass x temp change x specific heat)

PV = nRT R = 0.0821 L atm / mol K

P1V1 = P2V2 K = oC + 273 1 atm = 760 mmHg T1 T2

STP = 0oC and 1 atmPtotal = P1 + P2 + P3 + …

@ STP 1 mol = 22.4 L

NA = 6.02 x 1023