From the stoichiometry of the reaction, 1.4 * 10^-3 g is produced.
<h3>What mass of water is produced?</h3>
The equation of the reaction is written as; CO2 + 2LiOH → Li2CO3 + H2O. This can help us to apply the principle of stoichiometry here.
Thus;
Number of moles of CO2 = 0.00345 g/44 g/mol = 7.8 * 10^-5 moles
If 1 mole of CO2 produced 1 mole of water
7.8 * 10^-5 moles of CO2 produced 7.8 * 10^-5 moles of water
Mass of water produced = 7.8 * 10^-5 moles * 18 g/mol = 1.4 * 10^-3 g
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