if 20 gms of CaCO3 is treated with 20 gms of hcl, how many grams of CO2 can be generated according to given equation: CaCO3 (s) +2HCl (aq) - CaCl2 (aq) +H2O (l) +CO2(g)

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Solution

Molecular mass of CaCO₃ = 100 g
Molecular mass of HCl = 36.5 g
100 g of CaCO₃ reacts with 73 g (36.5 x 2) of HCl to give 44 gram of CO₂ according to given equation.
20 gram of CaCO₃will react with 73/100 *20 g HCl = 14.6 g. HCl only to give CO₂. So the limiting reagent is CaCO₃.
Since 100 g CaCO₃ give 44 g CO₂
**therefore 20 g of CaCO₃ will give = 44/100 *20 = 8.8 g CO₂.**

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Calcium carbonate reacts with aqueous $HCl$ to give ${CaCl}_{2}$ and ${CO}_{2}$ according to the reaction given below

$\underset{Calcium Carbonate}{{CaCO}_{3} (s)} + \underset{Hydrochloric acid}{2 HCl (aq)} ⟶ \underset{Calcium Chloride}{{CaCl}_{2} (aq)} + \underset{Carbon dioxide}{{CO}_{2} (g)} + \underset{Water}{H_{2} O (l)}$

What mass of ${CaCO}_{3}$ is required to react completely with 25 mL of 0.75 M HCl?

1) 0.6844

2) 0.9375

3) 0.4265

4) 0.2785

Calcium carbonate reacts with aqueous HCl to give CaCl₂and CO₂according to the reaction, CaCO₃+ 2HCl --> CaCl₂+ H₂O

What mass of CaCO3 is required to react completely with 25mL of 0.75 HCl?

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