Table of Contents
- 1 How much volume of carbon dioxide is produced when 50g of calcium carbonate is heated?
- 2 What is the volume occupied by CO2 gas after decomposition of 500g of CaCO3?
- 3 How much volume of CO2 is produced when 50g of calcium carbonate is heated completely under standard conditions CaCO3 CaO +CO2?
- 4 What will be the volume of CO2 produced by heating 0.25 g of CaCO3?
- 5 When 1g of CaCO3 reacts with 50ml of 0.1 M HCl the volume of CO2 produced is?
- 6 What volume of CO2 is liberated at NTP from 0.1 mole of CaCO3?
How much volume of carbon dioxide is produced when 50g of calcium carbonate is heated?
Hence, at STP, ${ 11.2L }$ of carbon dioxide is obtained by thermal decomposition of ${ 50g }$ of calcium carbonate.
What is the volume occupied by CO2 gas after decomposition of 500g of CaCO3?
0.5 mol * 22.4 L/mol = 11.2 L.
What volume of CO2 gas is produced at STP when 50 gm of CaCO3 reacts with excess of diluted HCl?
= 2.24 L of CO2 is liberated at STP.
What volume of CO2 is obtained at STP?
At STP: 1 mole of a gas occupies 22.4 L of volume. So, 0.093 moles of carbon dioxide will occupy = 0.093X22. 4=2.08 L of carbon dioxide.
How much volume of CO2 is produced when 50g of calcium carbonate is heated completely under standard conditions CaCO3 CaO +CO2?
Calcium carbonate decomposes on heating according to the equation; CaCO3(s)→CaO(s)+CO2(g) At STP, the volume of carbon dioxide obtained by thermal decomposition of 50g of calcium carbonate will be: (A) 22.4litre.
What will be the volume of CO2 produced by heating 0.25 g of CaCO3?
Therefore 25 g of CaCO3 gives 44/100 * 25 = 11 grams. One mole of CO2 is 44 grams. Therefore in 11 grams of CO2 , there would be 1/44 * 11 = 0.25 moles. 5.6 L of volume.
What volume of CO2 is formed at STP when 10 g of CaCO3 undergoes calcination?
10 g calcium carbonate corresponds to 0.1 mol. It on decomposition will give 0.1 mol of carbon dioxide. But limestone is only 90\% pure. Hence, 0.09 mol of carbon dioxide will be obtained.
What volume of CO2 gas is obtained by the decomposition of 50gm of CaCO3 at 273 C?
22 g
22 g of CO2 are obtained from decomposition of 50 g of CaCO3.
When 1g of CaCO3 reacts with 50ml of 0.1 M HCl the volume of CO2 produced is?
°• Volume of produced is 56ml.
What volume of CO2 is liberated at NTP from 0.1 mole of CaCO3?
CaCO3 +2HCl===> CaCl2 + CO2+ H2O. So 100 g of CaCO3 gives= 1 mole CO2. Volume of 1 mole of CO2 at NTP =22400ml. So volume of 0.01 mole of CO2 at NTP= 224 ml.
What volume of CO2 at STP can be obtained on the decomposition of 100.0 g of CaCO3?
or it can be represented as 100 grams when it is expressed as molecular mass. CO2 molar mass is 44 u, it’s molecular mass is 44 grams. This implies 100 g of CaCO3 gives 44 g of CO2.
What volume of CO2 is obtained at STP by heating of 4 grams of CaCO3?
4×22.4100 = 0.896 lit.