With 100 mL, 0.1M [tex]Na_2CO_3[/tex], the volume of 1M [tex]MgSO_4[/tex] that would be needed for complete reaction will be 0.01 L or 10 mL
Stoichiometric calculations
The equation of the reaction is as follows:
[tex]Na_2CO_3 + MgSO_4 ---- > Na_2SO_4 + MgCO_3[/tex]
Both reactants are in the ratio 1:1.
Mole = molarity x volume
Mole of 100mL, 0.1M [tex]Na_2CO_3[/tex] = 0.1 x 0.1 = 0.01 moles
Equivalent mole of [tex]MgSO_4[/tex] will also be 0.01 moles.
Volume of 0.01 moles, 1M [tex]MgSO_4[/tex] = 0.01/1 = 0.01 L or 10 mL
More stoichiometric calculations can be found here: https://brainly.com/question/27287858
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