1.How many moles of sodium oxide are produced when3.9 moles of sodium combine with excess oxygen.
2. Sodium chloride decomposes into elemental sodium and elemental chlorine.How many grams of chlorine will be produced if 39.4 grams of sodium chloride decomposes
3.If 6.7 moles of Mg react with 32 moles of O2.what is the limiting reactant
4.If 80 grams of sulfur dioxide reacts with 36 grams of carbon disulfide will be produced
5.If 84.8 grams of iron(III) oxide produced 57.8 grams of iron
4 answers
All of these are stoichiometry problems and all follow the same rules. What is your major hang up? For example, explain in detail what you don't understand about #1 and how to go about solving it.
I was able to solve the frist one.Some of problems come when i am finding the limiting reactant and decompisition.
#2.The problem TELLS you the decomposition products; i.e.,
2NaCl ==> 2Na + Cl2
Now you have a problem just like #1.
#3.
2Mg + O2 ==> 2MgO
Use the coefficients in the balanced equation to convert mol of EACH to mols MgO (the product)
6.7 mols Mg x (2 mols MgO/2 mol Mg) = 6.7 x 2/2 = 6.7 mols MgO formed IF we used the Mg in the problem and all of the O2 we needed.
32 mols O2 x (2 mols MgO)/1 mol O2) = 32 x 2/1 = 64 mols MgO IF we used 32 mols O2 and all of the Mg we needed.
Of course BOTH of these answers can't be right; one must be wrong. In limiting reagent problems the correct answer is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Thus Mg is the limiting reagent because it produces the smaller number of mols of product.
#4. This doesn't appear to be a complete question; however, if you complete the part that is omitted, then it is another limiting reagent problem but it strts with grams instead of mols. To convert to mols, mol = grams/molar mass and from there it's just like #3.
#5.
This is not a complete question.
2NaCl ==> 2Na + Cl2
Now you have a problem just like #1.
#3.
2Mg + O2 ==> 2MgO
Use the coefficients in the balanced equation to convert mol of EACH to mols MgO (the product)
6.7 mols Mg x (2 mols MgO/2 mol Mg) = 6.7 x 2/2 = 6.7 mols MgO formed IF we used the Mg in the problem and all of the O2 we needed.
32 mols O2 x (2 mols MgO)/1 mol O2) = 32 x 2/1 = 64 mols MgO IF we used 32 mols O2 and all of the Mg we needed.
Of course BOTH of these answers can't be right; one must be wrong. In limiting reagent problems the correct answer is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Thus Mg is the limiting reagent because it produces the smaller number of mols of product.
#4. This doesn't appear to be a complete question; however, if you complete the part that is omitted, then it is another limiting reagent problem but it strts with grams instead of mols. To convert to mols, mol = grams/molar mass and from there it's just like #3.
#5.
This is not a complete question.
buytj
1 answer