Answer:
To find the mass of lithium carbonate (\(Li_2CO_3\)) produced, we first need to calculate the moles of \(Li_2CO_3\) formed using the stoichiometry of the given chemical equation.
The equation provided is:
\[CO_2(g) + 2LiOH(s) \rightarrow Li_2CO_3(s) + H_2O(l)\]
This equation shows that 1 mole of \(CO_2\) reacts with 2 moles of \(LiOH\) to produce 1 mole of \(Li_2CO_3\) and 1 mole of \(H_2O\). Thus, the mole ratio of \(CO_2\) to \(Li_2CO_3\) is 1:1.
If 20.0 moles of \(CO_2\) is used:
- 20.0 moles of \(CO_2\) will produce 20.0 moles of \(Li_2CO_3\), given the 1:1 mole ratio from the equation.
Next, we calculate the mass of \(Li_2CO_3\) produced using its molar mass. The molar mass of \(Li_2CO_3\) can be calculated as follows:
- Molar mass of Li = 6.94 g/mol
- Molar mass of C = 12.01 g/mol
- Molar mass of O = 16.00 g/mol
So, the molar mass of \(Li_2CO_3 = 2(6.94) + 12.01 + 3(16.00)\) g/mol
\[= 13.88 + 12.01 + 48.00 \, \text{g/mol}\]
\[= 73.89 \, \text{g/mol}\]
Now, using the molar mass of \(Li_2CO_3\) to find the mass produced from 20.0 moles:
Mass of \(Li_2CO_3 = \text{moles} \times \text{molar mass}\)
\[= 20.0 \, \text{moles} \times 73.89 \, \text{g/mol}\]
\[= 1477.8 \, \text{g}\]
Therefore, 1477.8 grams of \(Li_2CO_3\) is produced
