The balance chemical equation is :
2HCL + Na2CO3 -----> 2NaCl + H2O + CO2
According to Question,
Given,
Molarity of HCL = 1.75 m
Sodium Carbonate (Na2CO3) = 0.100 m
For finding out the molarity,
c = n solute / Vsolution => n = c.Vsolution
nCO^2-3 = 0.100 mol L^-1 . 0.750 L = 0.0750 moles of CO^2-3
0.0750 moles of CO^2-3 . 2 moles H3O^+ / 1 mole CO^2-3
= 0.150 moles of H3O^+
As we have already know the molarity of HCL , we easily calculate what volumes by many moles.
c = nsolute / Vsolution => Vsolution = nsolute / c
VH3O^+ = 0.150 moles / 1.75 mol L^-1 = 0.0857 L
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