When solving for the answer to the problem, the following pattern is used. The missing information is given letter designations A through F, for labeling purposes. LO 3.45x1021 fu Cr(CIO3)3 1 X Х mol C mol D X Х E F​

23.5

23.5

47.0

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

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.