Testing on a water sample shows the following analysis. Calculate the total, carbonate and non-carbonate hardness in mg/L as CaCO3, and construct a bar chart (in meq/L) to represent the composition of the water.
Concentration Concentration
Chemical species (mg/L) Chemical species (mg/L)
Na⁺ 14.0 SiO2 3.4
Ca⁺² 96.8 F⁻ 1.1
Mg⁺² 30.4 Cl⁻ 4.0
Fe⁺³ 0.5 SO4⁻² 52.0
Mn⁺² 0.07 NO3⁻ 0.0
Zn⁺² 0.27 Alkalinity 260.7 as CaCO3
K⁺ 1.6 pH 7.7
Ba⁺² 0.2

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