Answer:
The mass, CO2 and CO3 from the limestone sample is discussed below in details.
Explanation:
(A) mass loss of sample of limestone after 20 min
= 0.8437g-0.5979g = 0.2458 g
From the given reaction of limestone, 2 mol of the sample gives 2 moles of CO 2.
Therefore
184.4 g ( molar mass of limestone) gives2× 44 g of carbon dioxide.
1 g of sample gives 88/184.4 g of carbon dioxide
Hence 0.2458 g sample gives
= 88/184.4 × 0.2458 g = 0.117 g carbon dioxide
(B) mole of CO 2 lost = weight/ molar mass
= 0.117 g / 44 g/mol =0.0027 mole
(C). 1 mol of limestone contain 2 mol of carbonate ion
From the reaction we know that carbonate ion of limestone is converted into carbondioxide
Hence lost carbonate ion = 0.2458 g
(D) we know that
1 mol limestone contain 1mol CaCO 3
Hence in sample present CaCO 3
= 1mole / 184.4 g × 0.8437 g= 0.00458 mol CaCO3
would be produced from the complete reaction of 25 mL of 0.833 mol/L 
with excess 
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