Can you tell me if these are right please?

When 2.50g of copper reacts with oxygen the copper oxide product has a mass of 2.81g. What is the simplest formula of the copper oxide.
I got Cu2O (copper (I) oxide).

Aluminum and oxygen gas react to produce aluminum oxide.

A.) write a balanced equation for the relation. I got 4Al+3O2=2Al2O3

B.) If 12 moles of Al2O3 are produced how many moles of aluminum reacted? I got 24 moles.

C.) if 75g of oxygen react how many grams of aluminum are required. I got 84.4 grams.

Using the rules for writing the formulas of ionic compounds, write the ions correct formula for magnesium oxide.

I got Mg2+O2= MgO is this right?

I agree with all of your answers but I would write the equation for the formation of MgO differently.

2Mg + O2 ==> 2MgO
MgO consists of Mg^+2 ion and O^-2 ion.

Sure, I'd be happy to help with your chemistry questions! Let's see how accurate your answers are:

1. Copper oxide: You got it right! The simplest formula for copper (I) oxide is Cu2O.

2. Aluminum and oxygen gas reaction:

A) Balanced equation: You are correct! The balanced equation for this reaction is 4Al + 3O2 → 2Al2O3.

B) Moles of aluminum: Good job! Since the balanced equation shows that 4 moles of aluminum react to produce 2 moles of Al2O3, if 12 moles of Al2O3 are produced, you would need 24 moles of aluminum.

C) Grams of aluminum: Well done! Given that 4 moles of aluminum react to produce 2 moles of Al2O3, you can use the molar mass of aluminum (26.98 g/mol) to calculate that you would need 84.4 grams of aluminum (24 moles x 26.98 g/mol).

3. Magnesium oxide formula: Unfortunately, you made a little mistake here. The correct formula for magnesium oxide is MgO. Magnesium is always found with a +2 charge, and oxygen is always found with a -2 charge. Therefore, to balance out the charges, you only need one magnesium ion (Mg2+) combined with one oxygen ion (O2-) to form MgO.

Great effort overall! Keep up the good work!

For the first question:

When 2.50g of copper reacts with oxygen, the copper oxide product has a mass of 2.81g. To find the simplest formula of the copper oxide, we need to calculate the mole ratios.

1. Convert the mass of copper to moles:
moles of copper = mass / molar mass
moles of copper = 2.50g / 63.55g/mol (molar mass of copper)
moles of copper = 0.0394 mol

2. Convert the mass of copper oxide to moles:
moles of copper oxide = mass / molar mass
moles of copper oxide = 2.81g / (63.55g/mol + molar mass of oxygen)
moles of copper oxide = 0.0442 mol

3. Divide both moles by the smallest value to determine the mole ratio:
mole ratio of copper to copper oxide = 0.0394 mol / 0.0394 mol (divided by itself)
mole ratio of copper to copper oxide = 1

mole ratio of oxygen to copper oxide = 0.0442 mol / 0.0394 mol
mole ratio of oxygen to copper oxide = 1.12

4. Simplify the mole ratios by dividing by the smallest whole number ratio (round as necessary):
mole ratio of copper to copper oxide ≈ 1
mole ratio of oxygen to copper oxide ≈ 1.12

Thus, the simplest formula of the copper oxide is CuO (copper (II) oxide) since the mole ratio of oxygen to copper oxide is closest to unity.

For the second question:

A.) The balanced equation for the reaction between aluminum and oxygen gas to produce aluminum oxide is:
4Al + 3O2 → 2Al2O3
Your balanced equation is correct.

B.) If 12 moles of Al2O3 are produced, we can use the mole ratio from the balanced equation to determine how many moles of aluminum reacted.
From the balanced equation, the mole ratio of aluminum oxide to aluminum is 2:4 or 1:2.
Moles of aluminum = 12 moles of Al2O3 * (2 moles of Al / 2 moles of Al2O3)
Moles of aluminum = 12 moles

Thus, 12 moles of aluminum reacted.

C.) If 75g of oxygen reacts, we need to determine how many grams of aluminum would be required.
From the balanced equation, the mole ratio of oxygen to aluminum is 3:4.
Moles of oxygen = 75g / 32.00g/mol (molar mass of oxygen)
Moles of oxygen = 2.34375 moles

Using the mole ratio, the moles of aluminum would be:
Moles of aluminum = 2.34375 moles of O2 * (4 moles of Al / 3 moles of O2)
Moles of aluminum = 3.125 moles

To convert moles of aluminum to grams:
Grams of aluminum = 3.125 moles of Al * 26.98g/mol (molar mass of aluminum)
Grams of aluminum = 84.41375g
Rounded to three significant figures, the answer is 84.4 grams.

For the last question:

To determine the correct formula for magnesium oxide using the rules for writing the formulas of ionic compounds, we need to consider the charges of the ions.

The ion charges for magnesium (Mg) and oxygen (O) are +2 and -2, respectively.

To write the formula, we need to balance the charges. Since the charges are equal in magnitude but opposite in sign, we only need one magnesium ion and one oxygen ion to balance.

Therefore, the correct formula for magnesium oxide is MgO. Your answer is correct.

Let's go through each question and check if your answers are correct:

1. The reaction of copper with oxygen produces copper oxide. Given that 2.50g of copper reacts to form 2.81g of copper oxide, we can calculate the ratio of the masses. The molar mass of copper is 63.55 g/mol. The molar mass of oxygen is 16.00 g/mol.

To determine the simplest formula of the copper oxide, we need to calculate the number of moles of copper and oxygen in the reaction.

Number of moles of copper = mass of copper / molar mass of copper
= 2.50g / 63.55 g/mol
≈ 0.039 moles

Number of moles of oxygen = mass of copper oxide - mass of copper / molar mass of oxygen
= (2.81g - 2.50g) / 16.00 g/mol
≈ 0.019 moles

Dividing the moles of copper and oxygen by the smallest value (0.019 moles) gives approximately 2:1 ratio, indicating that the simplest formula of the copper oxide is Cu2O (copper(I) oxide). So your answer is correct.

2. Balanced equation for the reaction between aluminum and oxygen to produce aluminum oxide:
4Al + 3O2 → 2Al2O3
Your answer is correct.

3. To determine the number of moles of aluminum reacted, we can use the stoichiometry of the balanced equation. From the equation, we know that 2 moles of aluminum react to produce 2 moles of Al2O3.

So, if 2 moles of Al2O3 are produced, then the number of moles of aluminum reacting would also be 2 moles. Therefore, your answer of 24 moles is incorrect.

4. If 12 moles of Al2O3 are produced, we know from the balanced equation that this corresponds to 2 moles of aluminum. So, the correct answer is 2 moles of aluminum.

5. To determine the grams of aluminum required when 75g of oxygen reacts, we can use the stoichiometry of the balanced equation. From the equation, we know that 3 moles of oxygen react to produce 2 moles of Al2O3.

First, calculate the number of moles of oxygen:
Number of moles of oxygen = mass of oxygen / molar mass of oxygen
= 75g / 16.00 g/mol
≈ 4.69 moles

From the stoichiometry of the balanced equation, we can see that 3 moles of oxygen react to produce 2 moles of Al2O3. So, we need to find the number of moles of Al2O3 produced by 4.69 moles of oxygen.

Number of moles of Al2O3 = (2 moles of Al2O3 / 3 moles of oxygen) * 4.69 moles of oxygen
≈ 3.13 moles

Finally, we can calculate the mass of aluminum required:
Mass of aluminum = number of moles of Al2O3 * molar mass of Al2O3
= 3.13 moles * (2 * molar mass of aluminum + 3 * molar mass of oxygen)
≈ 84.05 grams

So, the correct answer is approximately 84.05 grams. Therefore, your answer of 84.4 grams is incorrect.

6. The formula for magnesium oxide can be determined using the charges of the ions. Magnesium typically forms a 2+ charge, while oxygen forms a 2- charge.

The rule for writing a formula for ionic compounds is to balance the charges. Since magnesium has a 2+ charge and oxygen has a 2- charge, the simplest formula for magnesium oxide would be MgO (magnesium oxide). Therefore, your answer is correct.