When 1.2 x 10^-3 mol of nitric acid react with excess sodium carbonate, how many moles of carbonic acid are produced?
Technically, none since H2CO3 decomposes as fast as it is formed. H2CO3 ==> H2O + CO2. However, ignoring that, then
2HNO3 + Na2CO3 ==> 2NaNO3 + H2CO3
The equation tells you that 2 mols HNO3 will produce 1 mol H2CO3. So 1.2E-3 mols will produce ......?
To find the number of moles of carbonic acid produced, we need to use the balanced chemical equation for the reaction between nitric acid (HNO3) and sodium carbonate (Na2CO3). The balanced equation is as follows:
2 HNO3 + Na2CO3 → CO2 + H2O + 2 NaNO3
According to the balanced equation, we can see that the molar ratio between nitric acid and carbonic acid is 2:1.
In this case, 1.2 x 10^-3 mol of nitric acid is given. To convert this to moles of carbonic acid, we can use the following calculation:
Number of moles of carbonic acid = (Number of moles of nitric acid) x (Molar ratio)
Number of moles of carbonic acid = 1.2 x 10^-3 mol x (1 mol carbonic acid / 2 mol nitric acid)
(Number of moles of carbonic acid = 6.0 x 10^-4 mol)
Therefore, when 1.2 x 10^-3 mol of nitric acid react with excess sodium carbonate, 6.0 x 10^-4 mol of carbonic acid is produced.