Mgcl2 has a formula weight of 95g. mol-1. how would you make 10 ml of a 0.1 m solution of mgcl2.

moles = conc x vol  

moles = 0.1 x 0.010 = 0.001 moles  

mass = moles x molar mass  

mass = 0.001 moles x 95 = 0.095 g

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

the answers would be

the first one and the second one