Based on the reaction shown above, what volume of 0.246 M HCl is needed to titrate a 200 mL of 0.048 M Al(OH)3 solution? mL

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

weight of any object is given by

w = m*g

where m is mass of the object and g is the acceleration due to gravity.

for earth the value of g is 9.8 m/s^2 so

w=117.6 n m=12 g=?

117.6=12g => g=9.8 m/s^2

which is the force of gravity on earth.