Respuesta :
Moles of Li2CO3 = 1.53/73.891 = 0.0207 mole
Since HCl is in excess, amount of CO2 will depend on the limiting reagent which is Li2CO3.
∴Moles of CO2 = Moles of Li2CO3 = 0.0207.
Since HCl is in excess, amount of CO2 will depend on the limiting reagent which is Li2CO3.
∴Moles of CO2 = Moles of Li2CO3 = 0.0207.
Answer: The moles of carbon dioxide produced is 0.021 moles.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of lithium carbonate = 1.53 g
Molar mass of lithium carbonate = 73.9 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of lithium carbonate}=\frac{1.53g}{73.9g/mol}=0.021mol[/tex]
For the given chemical reaction:
[tex]Li_2CO_3(s)+2HCl(aq.)\rightarrow 2LiCl(aq.)+H_2O(l)+CO_2(g)[/tex]
As, hydrochloric acid is present in excess. So, it is considered as an excess reagent.
Lithium carbonate is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of lithium carbonate produces 1 mole of carbon dioxide
So, 0.021 moles of lithium carbonate will produce = [tex]\frac{1}{1}\times 0.021=0.021mol[/tex] of carbon dioxide
Hence, the moles of carbon dioxide produced is 0.021 moles.
