Setting up a chem problem
posted by Briana .
Magnesium reacts with water in the equation, Mg+2H²O-->Mg(OH)²+H². You have 8.1 g of magnesium available. This means you will need ___ g of water to use all the magnesium.
This seems sorta simple, but how can i set it up?
These are stoichiometry problems. They can be solved with a 4 step procedure. Remember this--it will solve many a problem for you.
Step 1. Write the balanced equation.
You have done that.
Mg + 2H2O ==> Mg(OH)2 + H2
Step 2. Convert what you have to mols remembering (for mass) that mols = grams/molar mass.
mols Mg = 8.1/24.3 = 0.333 mols.
Step 3. Using the coefficients in the balanced chemical equation, convert mols of what we have (in this case mols Mg) to mols of what we want (in this case H2O).
mols H2O = mols Mg x (2 mols H2O/1 mol Mg) = 0.333 x 2/1 = 0.667 mols H2O. [Note that the 2 mols H2O/1 mol Mg come directly from the equation.
Step 4. Now convert what we have from step 3 to grams.
grams = mols x molar mass
g H2O = mols H2O x molar mass H2O.
g H2O = 0.667 x 18 = ??
Check my work. Check my numbers. I estimated the molar masses so you will need to redo them and recalculate everything.
ok im working on it- slowly..
Briana--I'm calling it a night. If you have some question about the procedure, post a new question and someone will take care of it for you. I'm not the only one answering chemistry questions.
so- the answer would be 12.006? if that's right then yay! i think im understanding these...
Thanks for all your help!
See above. Yes, your answer is correct.
your answer is correct. Actually, two significant figures is the limit (since 8.1 has two s.f.) so 12 would be the correct answer for s.f. purists.