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slamgirl [31]
4 years ago
12

Calcium carbonate, when heated, forms calcium oxide and carbon dioxide. 100grams of calcium carbonate will produce 56grams of ca

lcium oxide. How many grams of carbon dioxide will it produce? Show working out
Chemistry
1 answer:
WITCHER [35]4 years ago
7 0

Answer:

It will produce 44 grams of carbon dioxide

Explanation:

Step 1: Data given

MAss of CaCO3 = 100 grams

Molar mass of CaCO3 = 100.09 g/mol

Mass of CaO produced = 56 grams

Molar mass of CaO = 56.08 g/mol

Step 2: The balanced equation

CaCO3 → CaO + CO2

Step 3: Calculate moles CaCO3

Moles CaCO3 = mass CaCO3 / molar mass CaCO3

Moles CaCO3 = 100 grams / 100.09 g/mol

Moles CaCO3 = 1.00 moles

Step 4: Calculate moles CaO

Moles CaO = 56 grams / 56.08 g/mol

Moles CaO = 1.00 moles

Step 5: Calculate moles CO2

For 1 mol CaCO3 we'll have 1 mol 1 mol CaO and 1 mol CO2

Step 6: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 1.00 moles * 44.0 g/mol

Mass CO2 = 44 grams

It will produce 44 grams of carbon dioxide

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Question 6 The mineral barite, (BaSO.) has a ke of 1.1 x 10" at 25°C. Calculate the solubility of barium sulfate in water, in: 6
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Explanation:

(6.1).    The reaction equation will be as follows.

             BaSO_{4}(s) \rightleftharpoons Ba^{2+}(aq) + SO^{2-}_{4}(aq)

Assuming the value of K_{sp} as 1.1 \times 10^{-10} and let the solubility of each specie involved in this reaction is "s". The expression for K_{sp} will be as follows.

            K_{sp} = [Ba^{2+}][SO^{-}_{2}]    (Solids are nor considered)

                        = s \times s

                   s = \sqrt{K_{sp}}

                      = \sqrt{1.1 \times 10^{-10}}

                      = 1.05 \times 10^{-5}

Therefore, solubility of barium sulfate in water is 1.05 \times 10^{-5}.

(6.2).   As the molar mass of BaSO_{4} is 233.38 g/mol

Therefore, the solubility is g/L will be calculated as follows.

                233.38 g/mol \times 1.05 \times 10^{-5}

                  = 2.45 \times 10^{-3} g/L

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3 years ago
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Read 2 more answers
How many grams of potassium iodine, kl,must be added to 500.0 grams of water to produce a 6.00% solution ?
GuDViN [60]

Answer:

grams KI needed = 31.9 grams

Explanation:

? g KI + 500 g water => 6.0% KI solution

let x = grams KI needed.

Percent = part/total x 100% = (x/x+500)·100% = 6.0%

Solve for 'x' ...

x / x + 500 = 6/100 = 0.06 => x = 0.06(x + 500) => x = 0.06x + 30

x - 0.06x = 30 => 0.94x = 30 => x = 30/0.94 = 31.9 gms KI needed.  

6 0
3 years ago
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