1.How many moles of sodium oxide are produced when3.9 moles of sodium combine with excess oxygen.

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.

buytj

1. To find the number of moles of sodium oxide produced, we need to determine the balanced chemical equation for the reaction between sodium and oxygen. The balanced equation is:

4 Na + O2 -> 2 Na2O

From the balanced equation, we can see that 4 moles of sodium react with 1 mole of oxygen to produce 2 moles of sodium oxide. Therefore, if 3.9 moles of sodium react, we can calculate the moles of sodium oxide produced:

Moles of sodium oxide = (3.9 moles of sodium) x (2 moles of sodium oxide / 4 moles of sodium) = 1.95 moles of sodium oxide

So, 1.95 moles of sodium oxide are produced when 3.9 moles of sodium combine with excess oxygen.

2. To determine the grams of chlorine produced when sodium chloride decomposes, we need to know the molar mass of sodium chloride and the balanced chemical equation for its decomposition. The molar mass of sodium chloride (NaCl) is 58.44 g/mol.

The balanced equation for the decomposition of sodium chloride is:

2 NaCl -> 2 Na + Cl2

From the balanced equation, we can see that 2 moles of sodium chloride decompose to produce 1 mole of chlorine gas. Therefore, we can calculate the moles of chlorine:

Moles of chlorine = (39.4 grams of sodium chloride) x (1 mole of sodium chloride / 58.44 grams of sodium chloride) x (1 mole of chlorine / 2 moles of sodium chloride) = 0.337 moles of chlorine

To find the grams of chlorine produced, we multiply the moles of chlorine by the molar mass of chlorine (Cl2):

Grams of chlorine = 0.337 moles of chlorine x (70.906 g/mol of chlorine) = 23.9 grams of chlorine

So, 23.9 grams of chlorine will be produced when 39.4 grams of sodium chloride decomposes.

3. To determine the limiting reactant between magnesium (Mg) and oxygen (O2), we need to compare the moles of each reactant and their stoichiometric ratio in the balanced chemical equation.

The balanced equation for the reaction between magnesium and oxygen is:

2 Mg + O2 -> 2 MgO

From the balanced equation, we can see that 2 moles of magnesium react with 1 mole of oxygen to produce 2 moles of magnesium oxide.

To determine the limiting reactant, we need to see which reactant is present in less quantity compared to their stoichiometric ratio.

For magnesium (Mg):
Moles of magnesium = 6.7 moles of Mg

For oxygen (O2):
Moles of oxygen = 32 moles of O2

Now, we can compare the moles of each reactant with their stoichiometric ratio:

Mg:O2 = 6.7 moles of Mg / 2 moles of Mg = 3.35
O2:Mg = 32 moles of O2 / 1 mole of O2 = 32

Since the ratio of Mg:O2 is 3.35, whereas the ratio of O2:Mg is 32, we can see that magnesium (Mg) is the limiting reactant, as it is present in less quantity compared to the stoichiometric ratio.

4. To determine the product formed when 80 grams of sulfur dioxide reacts with 36 grams of carbon disulfide, we need to determine the balanced chemical equation and the stoichiometry of the reaction.

Unfortunately, you didn't mention the specific reaction between sulfur dioxide and carbon disulfide, so I can't provide an accurate answer without that information. However, once you have the balanced chemical equation, you can calculate the moles of reactants and compare them to determine the limiting reactant and hence the product formed.

5. To determine the grams of iron (Fe) produced from a given mass of iron(III) oxide (Fe2O3), we need to know the molar mass of iron (Fe) and the balanced chemical equation for the reaction.

The balanced equation for the reaction between iron(III) oxide and iron is:

2 Fe2O3 -> 4 Fe + 3 O2

From the balanced equation, we can see that 2 moles of iron(III) oxide react to produce 4 moles of iron. Therefore, we can calculate the moles of iron:

Moles of iron = (84.8 grams of iron(III) oxide) x (1 mole of iron(III) oxide / molar mass of iron(III) oxide) x (4 moles of iron / 2 moles of iron(III) oxide)

Once you have the moles of iron, you can calculate the grams of iron using the molar mass of iron.

Please note that I can't calculate the exact answer for grams of iron without knowing the molar mass of iron(III) oxide.