Abeaker has 0.2 m of na2so4. what will be the concentration of sodium and sulfate ions?
[na+] = 0.4 m, [so42-] = 0.2 m
[na+] = 0.2 m, [so42-] = 0.2 m
[na+] = 0.4 m, [so42-] = 0.4 m
[na+] = 0.2 m, [so42-] = 0.4 m

The concentration of sodium and sulphate ions are [] = 0.4 M, [] = 0.2 M

Explanation:

The molar concentration is defined as the number of moles of a molecule or an ion in 1 liter of a solution.

In the given solution, the concentration of the salt sodium sulphate is 0.2M. So, 0.2 moles of sodium sulphate is present in 1 liter of solution.

Assuming 100% dissociation,

1 molecule of sodium sulphate gives 2 ions of sodium and 1 ion of sulphate.

So 0.2 moles of sodium sulphate will give 0.4 moles of sodium ions and 0.2 moles of sulphate ions.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2