25.0 cm3 of a solution of sodium hydroxide required 18.8 cm3 of 0.0500 mol/dm3 H2SO4

H2SO4+2NaOH—> Na2SO4+ 2H2O

1) find the concentration of the sodium hydroxide solution in mol/dm-3

2) find the concentration of sodium hydroxide solution in g/dm3

i think so

Ba(OH)2 + H2SO4 > BaSO4 + 2H2O

1) Moles of Ba(OH)2 = moles of H2SO4 = 0.025L x 2)0.02M = 5.0 x 10^-4M

Concn of Ba(OH)2 in g/L = 5.0 x 10^-4M x 171.33g/mol = 0.086g/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

lol idk man i guess u gotta ask bill nye the science guy