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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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Answer:

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In a first-order decay, the formula for the amount remaining after n half-lives is

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1. Calculate the number of half-lives.

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A dark brown binary compound contains oxygen and a metal. It is 13.38% oxygen by mass. Heating it moderately drives off some of

Leto [7]

Answer:

a) Mass of O in compound A = 32.72 g

Mass of O in compound B = 21.26 g

Mass of O in compound C = 15.94 g

b) Compound A = MO2

Compound B = M3O4

Compound C = MO

c) M = Pb

Explanation:

Step 1: Data given

A binairy compound contains oxygen (O) and metal (M)

⇒ 13.38 % O

⇒ 100 - 13.38 = 86.62 % M

After heating we get another binairy compound

⇒ 9.334 % O

⇒ 100 - 9.334 = 90.666 % M

After heating we get another binairy compound

⇒ 7.168 % O

⇒ 100 - 7.168 = 92.832 % M

The first compound has an empirical formula of MO2

⇒ 1 mol M for 2 moles O

Step 2: Calculate amount of metal and oxygen in each

compound A: M = m1 *0.8662 O = m1 *0.1338

compound B: M = m2 *0.90666 O = m2 *0.09334

compound C: M = m3 *0.92832 O = m3 *0.07168

Step 3: Calculate mass of oxygen with 1.000 grams of M

Compound A: 1.000g * 0.1338 m1gO / 0.8662m1gMetal = 0.1545

Compound B: 1.000g * 0.09334 m2gO / 0.90666m2gMetal = 0.1029

Compound C: 1.000g * 0.07168 m3gO / 0.92832m3gMetal = 0.07721

Step 4:

1 mol MO2 has 1 mol M and 2 moles O

m1 = (mol O * 16)/0.1338 m1 = 239.2 grams

1 mol M = 0.8632*239.2 = 206.48

0.90666m2 = 206.48 ⇒ m2 = 227.74 g

0.92832m3 = 206.48 ⇒ m3 = 222.42 g

Step 5: Calculate mass of O

Mass of O in compound A = 239.2 - 206.48 = 32.72 g

Mass of O in compound B = 227.74 - 206.48 = 21.26 g

Mass of O in compound C = 222.42- 206.48 = 15.94 g

Step 6: Calculate moles

Moles of O in compound A ≈ 2

⇒ MO2

Moles of O in compound B = 21.26 / 16 ≈ 1.33

⇒ M3O4

Moles of O compound C = 15.94 /16 ≈ 1 moles

⇒ MO

Step 7: Calculate molar mass

The mass of 1 mol metal is 206.48 grams ⇒ molar mass ≈ 206.48 g/mol

The closest metal to this molar mass is lead (Pb)

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