1) what mass of oxygen will react with 2.64g of magnesium?
2Mg + O2 -> MgO
I was going to do 2.64g * (1/2) from the equation but im not sure. Do I have to add a 2 in front of the MgO?
2) How many moles of sodium ions are present in 2.50 L of 0.300 M Na3PO4?
I know M=moles/L so .300= x/2.50 but the answer is 2.25 and i did not get that.
1) To find the mass of oxygen that will react with 2.64g of magnesium, you can use the stoichiometry of the balanced equation. In this case, the coefficient of magnesium (Mg) is 2, indicating that 2 moles of magnesium will react with 1 mole of oxygen (O2) to produce 2 moles of magnesium oxide (MgO).
Start by converting the mass of magnesium to moles using its molar mass. The molar mass of magnesium is 24.31 g/mol, so 2.64g of magnesium corresponds to:
2.64g * (1 mol/24.31g) = 0.109 mol of magnesium
Since the stoichiometric ratio between magnesium and oxygen is 2:1, you need twice as many moles of oxygen. Therefore, the number of moles of oxygen would be:
0.109 mol * (2 mol O2/2 mol Mg) = 0.109 mol of oxygen
Now, to find the mass of oxygen, multiply the number of moles of oxygen by its molar mass. The molar mass of oxygen is 31.998 g/mol, so:
0.109 mol * (31.998 g/mol) = 3.48 g of oxygen
Therefore, approximately 3.48 grams of oxygen will react with 2.64g of magnesium.
2) To determine the number of moles of sodium ions present in 2.50 L of 0.300 M Na3PO4, you can use the formula: moles = molarity * volume.
Given:
Molarity (M) = 0.300 M
Volume (V) = 2.50 L
Multiply the molarity by the volume to find the number of moles of Na3PO4:
0.300 M * 2.50 L = 0.75 moles of Na3PO4
In the balanced equation for Na3PO4, the coefficient of sodium ions (Na+) is 3. This means that 1 mole of Na3PO4 will yield 3 moles of sodium ions. Therefore, the number of moles of sodium ions present in 0.75 moles of Na3PO4 is:
0.75 moles Na3PO4 * (3 moles Na+/1 mole Na3PO4) = 2.25 moles of Na+
So, there are 2.25 moles of sodium ions present in 2.50 L of 0.300 M Na3PO4.