If 270. mL of 0.08089 M aqueous Pb(NO3)2 and 710. mL of 0.05493 M aqueous Na+ are reacted stoichiometrically according to the equation, what mass (g) of Na+ remained?
Pb(NO3)2(aq) + 2 NaI(aq) → PbI2(s) + 2 NaNO3(aq)
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You see how that works from the CaCO3 and H2SO4 problem just above so I think you can do this on your own. ust follow the steps. Post your work if you get stuck.
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