In the reaction MgCl2 + 2KOH → Mg(OH)2 + 2KCI, if 3 moles MgCl2 are added to 4 moles KOH, what determines how much Mg(OH)2 is made? A. The amount of KCI
B. The amount of MgCl2
C. The amount of MgCl2 + KOH
D. The amount of KOH

d. temperate deciduous forest

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

it is the most commonly accepted one