How many equivalents are present in 5.0 g of al3+?

The one definition of an equivalent weight that I know is that it is the amount of electrons which a substance gains or loses per mole.

Al3+ has +3 charge therefore it loses 3 e-, so we can say that there are 3 equivalent weights in 1 mol Al3+. 
1 mol Al3+ = 3 eq. wts. 
1 mol Al x (27 g / 1 mol ) x (1 mol / 3 eq.wts.) = 9.0 g = 1 eq.wt. 

5.0 g Al3+ x (1 eq.wt. / 9.0 g) = 0.56 eq.wts.

The number of equivalents present in 5 g of are 0.56.

Equivalent weight is the number of electrons gained or lost per mole by a material.

As has a +3 charge, it loses 3 electrons, resulting in 3 equivalent weights in 1 mol .

1 mole = 3 equivalent weight

1 mole Al = 27 g

Number of equivalent =

Number of equivalent = 9 in 1 mole.

27 grams = 9 equivalents

5 grams = 0.56 equivalent.

Thus, 5 grams of has been 0.56 equivalent.

To learn more about Equivalent weight, refer to the link:

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explanation:

i did this but i forgot how to do it

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