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3 years ago

How many milliliters of a 0.8 M solution of citric acid would be needed to react with 15 grams of baking soda? Show your work.

Chemistry

1 answer:

Alinara [238K]3 years ago

5 0

Answer:

75mL

Explanation:

Step 1:

The balanced equation for the reaction. This is illustrated below:

C6H8O7 + 3NaHCO3 → Na3C6H5O7 + 3CO2 + 3H2O

Step 2:

Conversion of 15g of baking soda (NaHCO3) to mole.

This is illustrated below:

Mass of NaHCO3 = 15g

Molar Mass of NaHCO3 = 23 + 1 + 12 + (16x3) = 23 + 1 + 12 + 48 = 84g/mol

Number of mole = Mass/Molar Mass

Number of mole of NaHCO3 = 15/84 = 0.179 mole

Step 3:

Determination of the number of mole of citric acid (C6H8O7) produced by the reaction. This is illustrated below:

From the balanced equation above,

1 mole of C6H8O7 reacted with 3 moles of NaHCO3.

Therefore, Xmol of C6H8O7 will react with 0.179 mole of NaHCO3 i.e

Xmol of C6H8O7 = 0.179/3

Xmol of C6H8O7 = 0.06 mole.

Step 4:

Determination of the volume of C6H8O7 needed for the reaction. This is illustrated below:

Molarity of C6H8O7 = 0.8 M

Mole of C6H8O7 = 0.06

Volume =..?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 0.06/0.8

Volume = 0.075L

Converting 0.075L to mL, we have:

1L = 1000mL

Therefore 0.075L = 0.075 x 1000 = 75mL.

Therefore, 75mL of citric acid (C6H8O7) is needed for the reaction.

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