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e-lub [12.9K]
3 years ago
6

Starting with 6.3 g of salicylic acid, how many grams of acetylsalicylic acid (theoretical yield) can be made? Assume salicylic

acid is the limiting reagent and a one to one stoichiometry. Use the molecular weights that you have already calculated. Give two significant digits.Answer:
Chemistry
1 answer:
tensa zangetsu [6.8K]3 years ago
4 0

Answer:

We will produce 8.2 grams of acetylsalicylic acid

Explanation:

step 1: Data given

Mass of salicylic acid, = 6.3 grams

Molar mass salicylic acid = 138.12 g/mol

Molar mass of acetylsalicylic acid = 180.158 g/mol

Step 2: The balanced equation

C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2

Step 3: Calculate moles salicylic acid

Moles salicylic acid = mass salicylic acid / molar mass salicylic acid

Moles salicylic acid = 6.3 grams / 138.12 g/mol

Moles salicylic acid = 0.0456 moles

Step 4: Calculate moles acetylsalicylic acid

Since the mole ratio is 1 to 1

For 0.0456 moles salicylic acid we'll have 0.0456 moles acetylsalicylic acid

Step 5: Calculate mass acetylsalicylic acid

MAss acetylsalicylic acid = moles acetylsalicylic acid * molar mass acetylsalicylic acid

MAss acetylsalicylic acid = 0.0456 moles * 180.158 g/mol

Mass acetylsalicylic acid = 8.2 grams

We will produce 8.2 grams of acetylsalicylic acid

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PLEASE HELP!!! NEED TO PASS THIS TO THE FIRST SEMESTER!
Cloud [144]

Answer:

Reaction 1 is balanced but 2 is not balanced , the balance equation are :

1. CH_{3}COOH(aq) + NaHCO_{3}(aq)\rightarrow CO_{2}(g) + H_{2}O(l) + CH_{3}COONa(aq)

2.CaCl_{2}(aq) + 2NaHCO_{3}(aq)\rightarrow CaCO_{3}(aq) + H_{2}O(l) + 2NaCl(aq) + H_{2}O(aq)

Explanation:

Balanced Equations : These are the equation which follows the law of conservation of mass .

The total number of atoms present in reactant is equal to total number of atoms present in product.

1. CH_{3}COOH(aq) + NaHCO_{3}(aq)\rightarrow CO_{2}(g) + H_{2}O(l) + CH_{3}COONa(aq)

This is acid - base type reaction where

CH_{3}COOH(aq) act as Acid

NaHCO_{3}(aq) act as weak base

Reactant :CH_{3}COOH(aq) ,NaHCO_{3}

Number of atoms of :

C = 2 (CH_{3}COOH(aq)) + 1 (NaHCO_{3})

   = 2 + 1

   = 3

H  = 4(CH_{3}COOH(aq)) + 1 (NaHCO_{3})

     = 4 + 1

       5

O = 2(CH_{3}COOH(aq)) + 3 (NaHCO_{3})

    = 5

Na = 1 (NaHCO_{3})

     = 1

Product :  CO_{2}(g),H_{2}O(l) , CH_{3}COONa(aq)

Number of atoms :

C = 1(CO_{2}(g)) + 2(CH_{3}COONa(aq))

   = 1 + 2

   = 3

H = 2(H_{2}O(l)) + 3(CH_{3}COONa(aq))

   = 2 + 3

   = 5

O = 1(H_{2}O(l)) + 2(CH_{3}COONa(aq))

        +2(CO_{2}(g)

    = 1 + 2 + 2

    = 5

Na = 1(CH_{3}COONa(aq)

     = 1

Number of Na =1 , C = 3 , H= 5 and O =5 in both reactant and product , so it is a balanced reaction

2.CaCl_{2}(aq) + 2NaHCO_{3}(aq)\rightarrow CaCO_{3}(aq) + H_{2}O(l) + 2NaCl(aq) + CO_{2}(g)

This is double displacement reaction .

Check the balancing in both reactant and products should be :

Na = 2

H = 2

Ca = 1

C = 2

O = 6

Cl = 2

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Polylactic acid is the correct answer
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A 1.20-L container contains 1.10 g of an unknown gas at STP. What is the molecular weight of the unknown gas?
SOVA2 [1]

Answer:

M = 20.5 g/mol

Explanation:

Given data:

Volume of gas = 1.20 L

Mass of gas = 1.10 g

Temperature and pressure = standard

Solution:

First of all we will calculate the density.

Formula:

d = mass/ volume

d = 1.10 g/ 1.20 L

d = 0.92 g/L

Now we will calculate the molar  mass.

d = PM/RT

0.92 g/L = 1 atm × M / 0.0821 atm.L/mol.K ×273.15 K

M =  0.92 g/L × 0.0821 atm.L/mol.K ×273.15 K /  1 atm

M = 20.5 g/mol

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Answer:

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there was leftover components meaning there was something mixed into the liquid

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