g The solution you created in this simulation was a 0.300M NH4Cl solution. The lab also stated that, in g/L, this concentration was 16.0 g/L. Show the calculations that prove that to be true.

16.0473 g/L

Explanation:

0.300 M=

0.300 mol/L x 53.491 grams/mol = 16.0473 grams/L

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