Assuming that you start with 2.00 g of sodium benzoate and 5 mL of 3 M...

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Assuming that you start with 2.00 g of sodium benzoate and 5 mLof 3 M HCl, how many molar equivalents of HCl are you using? Usetwo decimal places.

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4.0 Ratings (586 Votes)

mass of sodium benzoate = 2.00 g

molar mass of sodium benzoate= 144.11 g/mol

moles = mass / molar mass

moles of sodium benzoate (C6H5COONa) = 2 / 144.11

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.0139

moles of HCl = molarity x volume / 1000

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 3 x 5 / 1000

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.015

C6H5COONa +Â Â Â HClÂ Â Â ---------------------->Â Â C6H5COOH + NaCl

Â Â Â 1 molÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1 mol

0.0139 mol Â Â Â Â Â Â 0.015 mol

here limiting reagent is C6H5COONa

so 0.0139 mol of HCl is used.

molar equivalents of HCl = 0.01Â Â Â Â Â Â Â Â Â Â Â Â Â Â (two decimal places)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

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Assuming that you start with 2.00 g of sodium benzoate and 5 mL of 3 M...

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90.2K

Verified Solution

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Chemistry

Assuming that you start with 2.00 g of sodium benzoate and 5 mLof 3 M HCl, how many molar equivalents of HCl are you using? Usetwo decimal places.

Answer & Explanation Solved by verified expert

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