In an experiment, a student needs 250.0 mL of a 0.100 M copper (II) chloride solution. A stock solution of 2.00 M copper (II) chloride is available. How much of the stock solution is needed

Answer : The volume of stock solution needed are, 12.5 mL

Explanation :

Formula used :

where,

are the initial molarity and volume of copper (II) chloride.

are the final molarity and volume of stock solution of copper (II) chloride.

We are given:

Putting values in above equation, we get:

Hence, the volume of stock solution needed are, 12.5 mL

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g