Hello, Emi,
2 NaOH(aq) + Co(NO₃)₂(aq) → Co(OH)₂ (s) + 2 NaNO₃(aq)
The equation is balanced so we can move directly to converting the reactants into moles. I'll assume you know the basics behind moles, but please ask if something is unclear.
Moles NaOH: 32.0 g/40.0g/mole = 0.800 moles NaOH
Moles Co(NO3)2 : The definion of M is moles per liter. So we can multiply the concentration by the number of liters to find moles: (0.700 liters)*(1.00 moles/liter) = 0.700 moles Co(NO3)2
The balanced equation says 2 moles of NaOH are required for every 1 mole of Co(NO3)2. So if we start with 0.800 moles of NaOH, we'd need 1.600 moles of Co(NO3)2 for the reaction to consume all of the nNaOH. We just found that we ponly have 0.7 mole of Co(NO3)2 . That means it is the limiting reagent. The reaction will stop once it is all consumed. Base the next calculations on the assumption the reaction will stop once the Co(NO3)2 is gone.
The balanced equation says we need 2 moles of NaOH for every 1 mole of Co(NO3)2. That's a molar ratio of 1/2. If the entire 0.800 moles of NaOH is consumed, it would react with half that many moles of Co(NO3)2 . (1/2)*(0.800 moles) = 0.400 moles Co(NO3)2. That would leave 0.3 moles of Co(NO3)2 unreacted.
Looking at the Co(OH)2 produced, we find that it, also, only results in 1 mole for every 2 moles of NaOH. Assuming all of the NaOH is reacted (0.800 moles), we would wind up with 1/2 that many moles of Co(OH)2.
Therefore, we'd expect 0.400 moles of Ca(OH)2. Convert that to grams by multiplying by the molar mass of Ca(OH)2.
(0.400 moles Ca(OH)2)*(92.93 g/mole) = 37.2 grams Co(OH)2
Please ask if there is a step that isn't clear,
Bob
