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What volume of 2.25M NaCl solution is required to react completely with 1.00L of 0.750M Pb(NO3) solution? Pb(NO3)2(aq)+2NaCl(aq)=PbCl2(s)+2NaNO3(aq)

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You need to find the volume of NaCl, based from formula \[[M]=\frac{mol}{V}\]
They have given some values for Pb(NO3) right, so using that formula, \[[M]=\frac{mol}{V}\] mol=MV n(Pb(NO3)) = (0.750)(1.00) = 0.750moles of Pb(NO3) According to the equation, Pb(NO3)2(aq)+2NaCl(aq)=PbCl2(s)+2NaNO3(aq) 1 Pb(NO3)2 is equivalent to 2 NaCl, so 0.750 times 2 = 1.5 moles of NaCl then use back the equation \[[M]=\frac{mol}{V}\] \[V=\frac{mol}{[M]}\] \[V=\frac{1.5}{2.25}=0.667 m^3\]
Ok thank you so much! You just cleared up the M to mol to V formula for me. I was getting stumped using M to get to other values. All I knew was that M=mol/L Thank you for your help.This was the last practice problem I couldn't figure out. I might actually be ready for this quiz today now!

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0.667 m3 is a lot!

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