How many moles of co2 can be produced by the complete reaction of 1.53 g of lithium carbonatewith excess hydrochloric acid (balanced chemical reaction is given below)? li2co3(s) + 2hcl(aq) --> 2licl(aq) + h2o(l) + co2(g?

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.

Answer Link

Аноним Аноним · Answer 2 · 2019-12-18T08:09:21+01:00

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.