Calcium hydroxide (“slaked lime”) and sulfuric acid react to produce calcium sulfate and water according to the following balanced equation: Ca(OH)2 (aq) + H2SO4 (aq) -> CaSO4 (s) + 2 H2O (l)

Our H2SO4 solution is 0.5 M and we end up titrating (adding) 50 mL to an excess amount of Ca(OH)2 in H2O. Using the molarity and volume, how much CaSO4 do we end up producing in grams?

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

explanation: no but i do not usually comment