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As given 'Na3PO4' it is impossible to tell, because you also need the mass of sodium phosphate. Then use the eq'n
moles = mass(g) / Mr
From the Periodic Table we can calculate the Mr.
Naa x 3 = 23 x 3 = 69
P x 1 = 31 x 1 = 31
O x 4 = 16 x 4 = 64
69 + 31 + 64 = 164
164 is the Relative molecular mass (Mr) of sodium phosphate.
However,
mass(g) = 1 mole X 164(Mr)
mass of 1 mole = 164 grams.
Similarly
mass(g) = 2 moles X 164(Mr)
mass of 2 moles = 328 grams.
et seq.,
So you need to know the mass in order to calculate the moles.
What else can I help you with?
How many grams of Na3PO4 will be needed to produce 425 mL of a solution that has a concentration of Na plus ions of 0.900 M?
Let us find moles first. Molarity = moles of solute/Liters of solution ( 750 ml = 0.750 Liters ) 0.375 M Na2SO4 = moles Na2SO4/0.750 Liters = 0.28125 moles Na2SO4 =================== 0.28125 moles Na2SO4 (142.05 grams/1 mole Na2SO4) = 39.95 grams Na2SO4 needed ---------------------------------------you do significant figures!
How many moles are in 10 grams in sodium phosphate?
To calculate the number of moles in 10 grams of sodium phosphate (Na3PO4), we first need to determine the molar mass of Na3PO4, which is approximately 164.0 g/mol. Then, we can use the formula: moles = mass / molar mass. Therefore, for 10 grams of sodium phosphate, there would be approximately 0.061 moles present.
What is the mass of 4.76 moles of Na3PO4?
To calculate the mass of 4.76 moles of Na3PO4, you first need to find the molar mass of Na3PO4. Na has a molar mass of 22.99 g/mol, P has a molar mass of 30.97 g/mol, and O has a molar mass of 16.00 g/mol. The molar mass of Na3PO4 is (322.99) + 30.97 + (416.00) = 163.94 g/mol. Multiply the molar mass by the number of moles: 163.94 g/mol * 4.76 mol = 780.94 grams.
What mass of the precipitate could be produced by adding 100.0ml of 0.887 m agno3 to a na3po4 solution assume that the sodium phosphate reactant is present in excess?
To find the mass of the precipitate that forms when 100.0mL of 0.887M AgNO3 is added to a Na3PO4 solution, you need to determine the limiting reactant. Since Na3PO4 is in excess, AgNO3 is the limiting reactant. Calculate the moles of AgNO3 using its molarity and volume, then use the mole ratio between AgNO3 and the precipitate to find the moles of the precipitate. Finally, convert the moles of the precipitate to mass using its molar mass.
What is the molarity of 1 mol of Na3Po4 dissolved in 2L water?
The molarity of 1 mol of Na3PO4 in 2 L of water is 0.5 M. This is calculated by dividing the number of moles of solute (1 mol) by the volume of solution in liters (2 L).
How many moles of sodium ions are present in 2.50 L of 0.300 M Na3PO4?
There are 0.75 moles of sodium ions present in 2.50 L of 0.300 M Na3PO4. Each formula unit of Na3PO4 has 3 sodium ions. So, for every mole of Na3PO4, there are 3 moles of sodium ions. Calculating the moles of sodium ions: 2.5 L * 0.300 mol/L * 3 mol Na+ / 1 mol Na3PO4 = 0.75 moles of Na+.
How many moles are in 2 Na3PO4 3 MgCl2--Mg3PO42 6NaCl?
12 moles
In the formula for sodium phosphate Na3PO4 how many moles of sodium are represented?
There are 3 moles of sodium represented in one mole of sodium phosphate (Na3PO4). This is because the subscript 3 in Na3PO4 indicates that there are 3 sodium ions for every molecule of sodium phosphate.
How many grams of Na3PO4 will be needed to produce 425 mL of a solution that has a concentration of Na plus ions of 0.900 M?
Let us find moles first. Molarity = moles of solute/Liters of solution ( 750 ml = 0.750 Liters ) 0.375 M Na2SO4 = moles Na2SO4/0.750 Liters = 0.28125 moles Na2SO4 =================== 0.28125 moles Na2SO4 (142.05 grams/1 mole Na2SO4) = 39.95 grams Na2SO4 needed ---------------------------------------you do significant figures!
How many moles in 3.6x10-10g sodium phosphate?
2,196.10-12 pmol of Na3PO4. (p is pico)
How many moles of silver nitrate would be required to react with 2.00 moles of sodium phosphate?
I assume double displacement reaction. Balanced equation. 3AgNO3 + Na3PO4 -> Ag3PO4 + 3NaNO3 2.00 moles sodium phosphate ( 3 moles AgNO3/1 mole Na3PO4) = 6.00 moles silver nitrate needed =========================
100ml of 0.1M NaOH 50ml of 0.2M BaCl2 or 75ml of 0.15M Na3PO4 are strong electrolytes.Which contains the largest number of ions?
To determine which solution contains the largest number of ions, we need to calculate the total moles of ions produced by each solution. NaOH (100ml of 0.1M): 0.1 moles/L × 0.1 L = 0.01 moles of NaOH → produces 0.01 moles Na⁺ and 0.01 moles OH⁻ = 0.02 moles of ions. BaCl2 (50ml of 0.2M): 0.2 moles/L × 0.05 L = 0.01 moles of BaCl2 → produces 0.01 moles Ba²⁺ and 0.02 moles Cl⁻ = 0.03 moles of ions. Na3PO4 (75ml of 0.15M): 0.15 moles/L × 0.075 L = 0.01125 moles of Na3PO4 → produces 0.03375 moles Na⁺ and 0.01125 moles PO4³⁻ = 0.045 moles of ions. Therefore, Na3PO4 contains the largest number of ions.
How many moles are in 10 grams in sodium phosphate?
To calculate the number of moles in 10 grams of sodium phosphate (Na3PO4), we first need to determine the molar mass of Na3PO4, which is approximately 164.0 g/mol. Then, we can use the formula: moles = mass / molar mass. Therefore, for 10 grams of sodium phosphate, there would be approximately 0.061 moles present.
What is the mass of 4.76 moles of Na3PO4?
To calculate the mass of 4.76 moles of Na3PO4, you first need to find the molar mass of Na3PO4. Na has a molar mass of 22.99 g/mol, P has a molar mass of 30.97 g/mol, and O has a molar mass of 16.00 g/mol. The molar mass of Na3PO4 is (322.99) + 30.97 + (416.00) = 163.94 g/mol. Multiply the molar mass by the number of moles: 163.94 g/mol * 4.76 mol = 780.94 grams.
Number of oxygen atoms in 2.35 moles sodium phosphate?
There are 7.05 moles of oxygen atoms in 2.35 moles of sodium phosphate, as there are 3 oxygen atoms in each formula unit of sodium phosphate (Na3PO4).
What mass of the precipitate could be produced by adding 100.0ml of 0.887 m agno3 to a na3po4 solution assume that the sodium phosphate reactant is present in excess?
To find the mass of the precipitate that forms when 100.0mL of 0.887M AgNO3 is added to a Na3PO4 solution, you need to determine the limiting reactant. Since Na3PO4 is in excess, AgNO3 is the limiting reactant. Calculate the moles of AgNO3 using its molarity and volume, then use the mole ratio between AgNO3 and the precipitate to find the moles of the precipitate. Finally, convert the moles of the precipitate to mass using its molar mass.
What is the molarity of 1 mol of Na3Po4 dissolved in 2L water?
The molarity of 1 mol of Na3PO4 in 2 L of water is 0.5 M. This is calculated by dividing the number of moles of solute (1 mol) by the volume of solution in liters (2 L).
