Editing ReactionStoichiometry
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<b>1.</b> Write the balanced equation (reaction). <b>2.</b> Convert the known information (e.g. mass) of the reactant to moles. <ul><b>A.</b> If you need help on determining the apporpriate equation, look at the graphic organizer.</ul> <ul><b>B. </b> Will modify this step when we get to the Limiting Reactant problems.</ul> <b>3.</b> Use the balance equation to set-up the right "molar ratio" to go from the known substance to the unknown substance. <b>4.</b> Convert from the moles of the unknown susbtandce to tue unknown information (e.g., are you looking for mass, volume, # of particles) <ul><b>A.</b> Again, if you need help on determining the appropriate equation, look at the Graphical Organizer. </ul> <b>Example:</b> Calculate the mass (in grams) of NH<sub>3 </sub>(ammonia) produced by the reaction of 5.4 grams of hydrogen with nitrogen. The balanced reaction equation is: 1 N<sub>2</sub> (g) + 3 H<sub>2</sub> (g) => 2 NH<sub>3</sub> (g) <b>Step 1:</b> Question (see balanced reaction above) gave is information <b>Step 2:</b> Convert the mass of hydrogen (known substance) to moles # moles = grams/molar mass Molar Mass of H<sub>2</sub>: (1 g)(2) = 2 g/mol <b>Step 3:</b> Figure out from the balanced reaction what the Molar Ratio is between hydrogen and ammonia. 2 mole NH<sub>3</sub> / 3 mole H<sub>2</sub> Now use this ratio to convert from known (hydrogen) to unknown (ammonia). (2.70 mole H<sub>2</sub>)(2 mole NH<sub>3</sub> / 3 mole H<sub>2</sub>) = 1.80 mole NH<sub>3</sub> <b>Step 4</b>: Convert from unkown (ammonia) moles to unknown information (mass). #mole = grams/molar mass Molar Mass of NH<sub>3</sub>: (14 g)(1) + (1 g)(3) = 17 g/mol 1.80 mole NH<sub>3</sub> = X / 17 g/mol NH<sub>3</sub> X = (1.80 mole NH<sub>3</sub>)(17 g/mol NH<sub>3</sub>) = 30.60 grams NH<sub>3</sub> X = 30.6 grams NH<sub>3</sub>
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