Assume that 25 ml of 0.440m nacl are added to 25 ml of 0.320m agno3.
how many moles of agcl precipitate is?

nacl + agno3 -> nano3 + agcl

The balanced chemical equation would be as follows:

NaCl + AgNO3 -> NaNO3 + AgCl

We are given the amounts of the reactants. We need to determine first which one is the limiting reactant. We do as follows:

0.0440 mol/L NaCl (.025 L) = 0.0011 mol NaCl > consumed completely and therefore the limiting reactant
0.320 mol/L AgNO3 (0.025 L) = 0.008 mol AgNO3

0.0011 mol NaCl ( 1 mol AgCl / 1 mol NaCl) = 0.0011 AgCl precipitate produced

Rethink and reinvent