150ml of 0.8M lead(II)nitrate is reacted with 250ml of 0.6M sodium chloride which reactant is limiting and which is excess? What is the mass of the products? What is the net ionic equation for this reaction?

Question

vincent-lyon.fr · Answer

Write the equation first.

anonymous · Answer

PbNO3 + NaCl --> PbCl + NaNO3

vincent-lyon.fr · Answer

No, because the lead ion is Pb$^{2+}$

anonymous · Answer

are you referring to the net ionic equation?

vincent-lyon.fr · Answer

Yes, but the fact that Pb bears two positive charges also implies a different formula for lead nitrate and for lead chloride.

anonymous · Answer

Pb+3NO3 + 3NaCl --> PbCl3 + 3NaNO3
is this right?

aaronq · Answer

you need to balance they charges:  $Pb^{2+} + 2NO_3^-\rightarrow Pb(NO_3)_2$

anonymous · Answer

3Pb + 6NO3 --> 3Pb(NO3)2

is this right?

aaronq · Answer

nah i was writing the association of the ions, showing the charges balanced, it should be $Pb(NO_3)_2$

anonymous · Answer

acevsnake, do you still need help?

anonymous · Answer

ok -

anonymous · Answer

alright give me a sec

anonymous · Answer

what equation do you have?

anonymous · Answer

Pb2++2NO−3→Pb(NO3)2

anonymous · Answer

that is you net ionic?

anonymous · Answer

ah ok

anonymous · Answer

lets start from the beginning, i'll walk you through the problem.

anonymous · Answer

ok

anonymous · Answer

so what is your balanced molecular equation?

anonymous · Answer

Pb(NO3)2 + NaCl → PbCl + Na(NO3)2

anonymous · Answer

My battery is dieing just incase I do not respond