Reduction half-reaction E∘ (V) Ag+(aq)+e−→Ag(s) 0.80
Cu2+(aq)+2e−→Cu(s) 0.34
Sn4+(aq)+4e−→Sn(s) 0.15
2H+(aq)+2e−→H2(g) 0
Ni2+(aq)+2e−→Ni(s) −0.26
Fe2+(aq)+2e−→Fe(s) −0.45
Zn2+(aq)+2e−→Zn(s) −0.76
Al3+(aq)+3e−→Al(s) −1.66
Mg2+(aq)+2e−→Mg(s) −2.37

Calculate the standard cell potential (E∘) for the reaction

X(s)+Y+(aq)→X+(aq)+Y(s)

if K = 3.62×10−3.

Express your answer to three significant figures and include the appropriate units.

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

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

hope this