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The solidification of a $40$ % suspension of $\ce{Mg(OH)2}$ in water is probably due to the action of $\ce{CO2}$ from the air, which reacts with $\ce{Mg(OH)2}$ producing $\ce{MgCO3}$ which is insoluble in water. Remember that $58~ g$ $\ce{Mg(OH)2}$ produces $84~ g$ $\ce{MgCO3}$ and $174~ g$ lansfordite $\ce{MgCO3⋅5H2O}$, as described in the Handbook of ...