Respuesta :
1) Balance the chemical equation
[tex]CH_3COOH+NaOH\rightarrow NaCH_3COO+H_2O[/tex]2) Moles of NaOH in the reaction
Convert mL into L
[tex]L=15.53mL\cdot\frac{1L}{1000mL}=0.01553L[/tex]Find moles of NaOH
[tex]M=\frac{\text{moles of solute}}{\text{liters of solution}}[/tex]Plug in values and solve for moles
[tex]0.088M=\frac{\text{moles of NaOH}}{0.01553L}[/tex][tex]\text{moles of NaOH=0.088M}\cdot0.01553L=0.00136664\text{ mol NaOH}[/tex]3) Moles of acetic acid that reacted with 0.00136664 mol NaOH
Molar ratio
1 mol NaOH: 1 mol CH3COOH
[tex]\text{mol CH}_3COOH=0.00136664\text{ mol NaOH}\cdot\frac{1molCH_3COOH}{1\text{ mol NaOH}}=0.00136664molCH_3COOH[/tex]4) Convert moles of CH3COOH into grams of CH3COOH
The molar mass of CH3COOH is 60.05 g/mol
[tex]gCH_3COOH=0.00136664molCH_3COOH\cdot\frac{60.05gCH_3COOH}{1molCH_3COOH}=0.0820667gCH_3COOH[/tex]The mass of acetic acid is 0.0820667g
5) Percent of acetic acid
[tex]\text{Mass}\%=\frac{\text{grams of solute}}{\text{grams of solution}}\cdot100[/tex][tex]\text{Mass}\%=\frac{0.0820667gCH_3COOH}{1.98\text{ g of solution}}\cdot100=4.14\%[/tex]The percent of acetic acid by mass is 4.14%.
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