Question

how much caco3 required to get 22.71 dm3 of co2 at stp ​

Answer 1

Answer:

this is ans ...................

Answer 2

Explanation:

ANSWER

ANSWERThe reaction is CaCO

ANSWERThe reaction is CaCO 3

ANSWERThe reaction is CaCO 3

ANSWERThe reaction is CaCO 3 →CaO+CO

ANSWERThe reaction is CaCO 3 →CaO+CO 2

ANSWERThe reaction is CaCO 3 →CaO+CO 2

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One mole of calcium carbonate forms one mole of carbon dioxide.

One mole of calcium carbonate forms one mole of carbon dioxide.At STP, 5.6 L of carbon dioxide corresponds to 0.25 moles.

One mole of calcium carbonate forms one mole of carbon dioxide.At STP, 5.6 L of carbon dioxide corresponds to 0.25 moles. They will be obtained from 0.25 moles of calcium carbonate.

One mole of calcium carbonate forms one mole of carbon dioxide.At STP, 5.6 L of carbon dioxide corresponds to 0.25 moles. They will be obtained from 0.25 moles of calcium carbonate.Molar mass of CaCO

One mole of calcium carbonate forms one mole of carbon dioxide.At STP, 5.6 L of carbon dioxide corresponds to 0.25 moles. They will be obtained from 0.25 moles of calcium carbonate.Molar mass of CaCO 3

One mole of calcium carbonate forms one mole of carbon dioxide.At STP, 5.6 L of carbon dioxide corresponds to 0.25 moles. They will be obtained from 0.25 moles of calcium carbonate.Molar mass of CaCO 3

One mole of calcium carbonate forms one mole of carbon dioxide.At STP, 5.6 L of carbon dioxide corresponds to 0.25 moles. They will be obtained from 0.25 moles of calcium carbonate.Molar mass of CaCO 3 =100g

One mole of calcium carbonate forms one mole of carbon dioxide.At STP, 5.6 L of carbon dioxide corresponds to 0.25 moles. They will be obtained from 0.25 moles of calcium carbonate.Molar mass of CaCO 3 =100g0.25 moles of calcium carbonate (molecular weight 100 g/mole) corresponds to 25 g.