Question

1) According to the activity series, which of these metals will react with most acids to produce H2 gas?

Ag, Ni, or K?

2) Calculate the molarity:

a) 0.750 mol of Na2S in 1.30L of solution. b) 32.7g of MgS in 773ml of solution.

3) What is the reduction half-reaction in Cl2+2Na ---> 2NaCl?

4) How many mL of 11.0 M HCL (aq) are needed to prepare 720.0ml of 1.00 M HCL (aq)?

5) 63.0 ml of a 1.50M solution is diluted to a total volume of 248ml. A 124ml portion of that solution is diluted by adding 161ml of water. What is the final concentration? Assume the volumes are additive.

Answer 1

1) According to the activity series metal that are listed above hydrogen in the series will react faster with acids then the ones listed below.

Thus, of the three metals, potassium would react the most with acid to produce H2 gas.

2) Molarity = moles/L of solution = g of solute/molar mass x L of solution

a) 0.750 mol of Na2S in 1.30 L of solution

molarity = 0.750/1.30 = 0.58 M

b) 32.7 g of MgS in 773 mL of solution

Molarity = 32.7/(56.38 x 0.773) = 0.75 M

3) Reduction hald cell would be,

Cl2 + 2e- ---> 2Cl-

4) We would use the formula,

M1V1 = M2V2

where,

M1 = 11.0 M HCl

V1 = unknown

M2 = 1.00 M HCl

V2 = 720 mL

feed values,

11 x V1 = 1 x 720

V1 = 65.45 mL of 11.0 M HCl (aq) would be reuired to prepare a 1 M 720 mL HCl solution.

5) Initial moles of solution = molarity x volume = 1.5 x 0.063 = 0.0945 mols

First dilution,

total volume of solution = 248 mL = 0.248 L

Molarity = 0.0945/0.248 = 0.381 M

Second dilution,

124 mL of 0.381 M solution diluted by adding 161 mL of water

moles of solution = 0.381 x 0.124 = 0.047 mols

total volume of solution = 124 + 161 = 285 mL = 0.285 L

final concentration of solution = 0.165 M