Provided with enough heat, magnesium and oxygen can undergo a synthesis reaction. How much magnesium oxide is produced if 50g of Mg is reacted with 50g of oxygen?

This is a limiting reagent problem. How do you know that? Because BOTH reactants are given instead of just one. Basically, one works two simple stoichiometry problems.

1, Write the equation.
2Mg + O2 ==> 2MgO

2. Convert 50 g Mg to moles. moles = grams/molar mass.
3. Convert 50 g oxygen to moles. Same procedure.
4. Using the coefficients in the balanced equation, convert moles Mg to moles MgO.
5. Same procedure, convert moles oxygen to moles MgO.
6. It is likely that the moles MgO formed from the two previous steps will not agree which means one of them is wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that number is the limiting regent.
7. If you want the answer in grams, convert mols to grams. g = mols MgO x molar mass MgO.

Hey thank u for ur answer ... but can u show me with steps from question 4 to 6 please because I didn't get it what u mean there and please can u tell me what the answer will be thank you .....

To determine how much magnesium oxide is produced, we need to use stoichiometry, which involves balancing the chemical equation and converting the given masses into moles.

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

2 Mg + O2 -> 2 MgO

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

Step 1: Convert the given masses into moles.
To do this, we need to calculate the number of moles for each substance using their respective molar masses. The molar masses are as follows:

- Mg (magnesium): 24.31 g/mol
- O (oxygen): 16.00 g/mol

Number of moles of magnesium (Mg):
50 g / 24.31 g/mol = 2.06 mol

Number of moles of oxygen (O2):
50 g / 32.00 g/mol = 1.56 mol

Step 2: Determine the limiting reactant.
The limiting reactant is the one that is completely used up in the reaction and determines the amount of product formed. To find the limiting reactant, we need to compare the moles of each reactant to their stoichiometric ratio in the balanced equation. In this case, we compare the moles of magnesium to oxygen:

Moles of Mg / Stoichiometric coefficient of Mg = 2.06 mol / 2 = 1.03 mol
Moles of O2 / Stoichiometric coefficient of O2 = 1.56 mol / 1 = 1.56 mol

Since 1.03 mol of magnesium is less than 1.56 mol of oxygen, magnesium is the limiting reactant.

Step 3: Calculate the moles of magnesium oxide produced.
From the stoichiometry of the balanced equation, we know that 2 moles of magnesium oxide (MgO) are produced for every 2 moles of magnesium. Therefore, the number of moles of magnesium oxide produced is equal to the number of moles of magnesium used.

Moles of MgO produced = Moles of Mg used = 1.03 mol

Step 4: Convert moles of magnesium oxide to grams.
To convert the moles of magnesium oxide to grams, we need to multiply the moles by the molar mass of magnesium oxide (MgO), which is 40.31 g/mol.

Grams of MgO produced = Moles of MgO * Molar mass of MgO
Grams of MgO produced = 1.03 mol * 40.31 g/mol = 41.53 g

Therefore, when 50 g of magnesium is reacted with 50 g of oxygen, the amount of magnesium oxide produced is approximately 41.53 g.