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1) How many grams of PbCl2 are formed when 25.0 mL of 0.654 M KCl react with Pb(NO3)2?
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!)
ANSWER= [NH3]2/[N2][H2]3
PLEASE EXPLAIN THIS AND HOW TO DO GET THAT?
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