Is all this right?

What mass of sodium sulfate is produced when 25.0 mL of 0.100 M sulfuric acid is added to 20.0 mL of 0.15 M sodium hydroxide solution?

[latex]n=cV[/latex]

[latex]n(H_{2}SO_{4})=0.100*0.0250=0.00250=2.50*10^{-3} mol[/latex]

[latex]n=cV[/latex]

[latex]n(NaOH)=0.15*0.0200=0.0030=3.0*10^{-3} mol[/latex]

[latex]\frac{n(H_{2}SO_{4})}{n(NaOH)} =\frac{1}{2}[/latex]

[latex]n(H_{2}SO_{4})[/latex] required [latex]= \frac{1}{2} * (3.0*10^{-3}) = 0.0015 = 1.5*10^{-3} mol[/latex]

[latex]n(H_{2}SO_{4})[/latex] is in excess by [latex](2.50*10^{-3}-1.5*10^{-3}) 1.0*10^{-3} mol[/latex]. NaOH is the limiting reagent.

[latex]\frac{n(Na_{2}SO_{4})}{n(NaOH)} = \frac{1}{2}[/latex]

[latex]n(Na_{2}SO_{4})=\frac{1}{2}*(3.0*10^{-3}) =0.0015=1.5*10^{-3} mol[/latex]

[latex]m=n*M[/latex]

[latex]m(Na_{2}SO_{4})=\frac{1}{2}*(1.5*10^{-3}) * (2*22.9898+32.064+15.9994*4)= (1.5*10^{-3})*142.0412[/latex]

[latex]=0.2130618[/latex]

[latex]=0.21 g[/latex]

Did I make any mistakes?

Edited March 30, 201114 yr by gerald_mcdonald