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In-s [12.5K]
2 years ago
6

Write an equation for the thermal decomposition of CaCO3. Determine the volume of CO2 measured at s t p. that would be produced

by the thermal decomposition of 10g of CaCO3. [Ca= 40, O =16 C= 12, molar volume = 22.4dm3 at STP]​
Chemistry
1 answer:
grin007 [14]2 years ago
8 0

Answer : 2.24 L / 2.24 dm³

We need to write a equation for the thermal decomposition of Calcium carbonate . The decomposition is as follows ,

\rm CaCO_3 \xrightarrow{\Delta} CaO + CO_2

The molecular mass of Calcium carbonate is ,

\rm\longrightarrow Molecular\ mass_{CaCO_3}= 40g + 12g + (16)(3)g = 100g

The reaction is already balanced.

  • From the reaction , 100g of Calcium carbonate gives 1 mole of carbon dioxide.

\text{ 100g of Calcium carbonate $\equiv$ 1 mole of carbon dioxide }\\\\\text{ 1g of calcium carbonate $\equiv$ $\rm \dfrac{1}{100}$ mol of carbon dioxide.}\\\\\text{ 10g of calcium carbonate$\equiv$ $\rm \dfrac{1}{100}\times 10= $ 0.1 mol of carbon dioxide.}

  • And at STP , we know that the volume of 1 mol of gas is 22.4L or 22.4 dm³ .

So that,

\longrightarrow\rm 0.1\  mole \ occupies \equiv 0.1(22.4L) = 2.24 L

Hence ,

\longrightarrow \underline{\boxed{\bf{\red{ Volume = 2.24 L = 2.24\ dm^3}}}}

I hope this helps .

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We calculate the moles of Fe⁺² from the used moles of KMnO₄:

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Then we substract those 0.30 mmol from the original amount used:

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The reaction between ferrous ammonium sulfate and MnO₂ is:

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So we convert those 3.5 mmol Fe⁺² that were used in this reaction to MnO₂ moles:

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Then we convert MnO₂ to Mn₃O₄, using the reaction:

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Finally we convert Mn₃O₄ moles to grams:

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And calculate the percent

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