Calculate the number of moles of NaOH contained in 250. mL of a 0.05M solution?

0.0125 moles of NaOH are present

Explanation:

Molarity, M, is an unit of concentration widely used in chemistry. Is defined as the ratio between moles of solute (In this case, NaOH) and liters of solution.

250.0mL are = 0.250L of solution. As the molarity of the solution is 0.05M = 0.05moles / 1L, the moles present are:

0.250L * (0.05moles / 1L) =

output?

explanation:

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