If 6.09 g of sodium chloride are mixed with 35.50 g of water, what is the mass % (w/w%) salt in the solution

14.6 %

Explanation:

Step 1: Given data

Mass of sodium chloride (solute): 6.09 g

Mass of water (solvent): 35.50 g

Step 2: Calculate the mass of solution

We will use the following expression.

m(solution) = m(solute) + m(solvent)

m(solution) = 6.09 g + 35.50 g

m(solution) = 41.59 g

Step 3: Calculate the percent by mass of the salt

We will use the following expression.

%w/w = mass of NaCl / mass of solution × 100%

%w/w = 6.09 g / 41.59 g × 100%

%w/w = 14.6 %

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