Solution for QID #4712: 1) How many grams of PbCl2 are formed when 25.0 mL of 0.654 | StudyHelpMe

1) How many grams of PbCl2 are formed when 25.0 mL of 0.654 M KCl react with Pb(NO3)2?  2KCl(aq) + Pb(NO3) 2(aq) → 2KNO3(aq) + PbCl2(s) Answer= 2.27 g I WAS SHOWN TO DO IT AS PRESENTED BELOW: PLEASE SHOW ME A SIMPLE, MORE STRAIGHT FORWARD WAY TO DO THE MATH, I AM NOT UNDERSTANDING THIS WAY. ALSO NOT SURE WHERE THE 10^3 CAME FROM?  moles of KCl= Molarity×volume in L Moles= 0.654M×(25.0/1000)L=0.01635 moles 2KCl(aq) + Pb(NO3) 2(aq) → 2KNO3(aq) + PbCl2(s) From the balanced equation, mole ratio of KCl: PbCl2=2:1 Moles of PbCl2= 0.01635 moles/2=8.175× 10-3moles Moles=mass in g/RFM RFM for PbCl2=278g/mol Mass in g=moles ×RFM Mass in g=8.175× 10-3moles×278g/mol (WHEN I CALC THIS IT EQUALS 22.7?) Mass in g=2.27g 2) The equilibrium constant for the formation of ammonia (NH3) from nitrogen and hydrogen is 1.6 × 102. What is the form of the equilibrium constant? (Hint: balance the reaction!)