10.0 points 3.1 mol Fe(OHI); and 6.9 mol IL, SO, react
according to the equation
2 Fe(OH)3 + 3H2SO,
Fe2(SO4)3 + 6H20.
If the limiting reactant is Fe(OH)2, deter-
mine the amount of excess reactant that re-
mains.
Answer in units of mol.

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

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year

un recipient conține 6,4 moli de gaz. gazul hidrogen reprezintă 25% din totalul kilometrilor din container. în cazul în care presiunea totală este de 1.24 atm ceea ce este