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olga nikolaevna [1]
3 years ago
12

If you have 8.5 moles of Al2(SO4)3 how many grams of the compound do you have?

Chemistry
1 answer:
Zepler [3.9K]3 years ago
5 0

Answer:

RFM \: of \: aluminium \: sulphate = (2 \times 27) + (32 + (16 \times 4)) \times 3 \\  = 294 \\ 1 \: mole \: weighs \: 294 \: g \\ 8.5 \: moles \: weigh \: (8.5 \times 294) \: g \\  = 2499 \: g

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3 years ago
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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