AgBr, a chemical used in photography, can be made with this reaction:
2AgNO3 + CaBr2 > 2 AgBr + CaN2O6
AgNO3 is very expensive, so making AgBr, you want to be sure that this reactant is all used up. If a chemist starts with 28.0g of AgNO3, how many grams of CaBr2 must be added to use up all of the AgNO3?
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2AgNO3 + CaBr2 --> 2AgBr + CaN2O6
We have 28 g of AgNO3. We can see from the above balanced chemical equation that for every two moles of AgNO3 we need one mole of CaBr2. So let's convert the mass of AgNO3 into moles.
28 g AgNO3 * (1 mole/169.87 g) = 0.165 moles AgNO3
We have stated that the molar ratio of AgNO3 to CaBr2 is 2:1 according to the balanced chemical equation. So we divide the moles of AgNO3 by 2 to get the number of moles of CaBr2 needed to completely react with it.
0.165 moles AgNO3 * (1 mole CaBr2/2 moles AgNO3) = 0.0825 moles CaBr2
Now convert the moles of CaBr2 into grams.
0.0825 moles CaBr2 * (199.89 g/mole) = 16.5 g CaBr2.
Since silver is very expensive, we need to make sure that it is the limiting reagent in the reaction (in other words that it is not in excess and therefore wasted). That means that we need to either have equimolar amounts of both starting reagents or have the CaBr2 be in excess. In order for that to be true, we need to have at least 16.5 g of CaBr2 present in the reaction.
To make sure that all the AgNO3 is used in the reaction, you should add CaBr2 in excess. Use stoichiometry to find the value of CaBr2 needed. The units should cancel out leaving "grams CaBr2."
`28.0 "gAgNO3" = (1 "molAgNO3") / (169.87 "gAgNO3") = (1"molCaBr2")/(2"molAgNO3") = (199.98"gCaBr2")/(1"molCaBr2") = 16.46"gCaBr2"`
Multiply the top values and then divide it with the product of the denominators to get your answer.
Thus, you would need more than 16.5 grams of CaBr2 (you need three sig figs!).
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