Lead (II) Nitrate + Sodium Bromide

Pb(NO3)2+NaBr- By: Gabriella Cortez

How to make a balanced equation:

To make a balanced equation for the double replacement reaction Pb(NO3)2 + NaBr, you need to switch the first two metals and then distribute charges. If the reactants have the same number of elements then they are balanced but if they don't, then you need to add coefficients before the element to balance on both sides. After both sides of the reactant are balanced then your whole equation is balanced leaving you with Pb(NO3)2+2NaBr-->2Na(NO3)+PbBr2

IUPAC name for each reactant and product:

Pb(NO3)2+NaBr---> Lead (II) Nitrate and sodium Bromide

How to calculate molar mass for Lead (II) Nitrate + Sodium Bromide

To find the molar mass of an equation or reaction you will need to use information located on the periodic table. To find what mass your element has, find it on the periodic table and look at the number below it. My elements were Pb, N, O, Na, and Br. After finding all of their masses you can multiply them by the number that they have after them, for example, O6, which means I will calculate the mass six times. After calculating each individual mass you can add them all up to get your molar mass. The molar mass for one side of my balanced equation is 311.208.

My mass to mass conversions:

Starting off in a conversion chart, the top left of mine was 12.1 grams of Pb(NO2)2 because I needed to find how many grams of NaBr that could go in. After starting off my conversion chart I put what I wanted to be cancelled out on the bottom to get a number of moles for reaction NaBr. Multiplying both top and bottom and then dividing my total was 7.5 grams NaBr.

Limiting and Excess reactant:

Limiting reactant problems are simply where one of the reactants runs out first, leaving more amounts of the other reactants called excess reactants.
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Theoretical & Percent yield:

Theoretical yields are the amount of products that are supposed to be created by a chemical reaction, if none of the reactants were wasted then the reaction was fully completed. The first thing to know is your limiting reactant, mine is 12.3 Pb(NO3)2. You know that your actual yield is by your mass to mass formula, mine was 7.23g of Na(NO3). Diving your actual yield by your theoretical yield and then multiplying by 100 is how you find your percent yield. I divided 7.23 by 6.31 (theoretical) then multiplying by 100 to get my percent yield of 144.5%.
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