Suppose a solution containing 3.50 g of Na3PO4 is mixed with an excess of Ba(NO3)2. How many grams of Ba(PO4)2 can be formed.
To find the number of grams of Ba(PO4)2 that can be formed, we need to determine the limiting reactant in the reaction between Na3PO4 and Ba(NO3)2. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Let's first write down the balanced chemical equation for the reaction between Na3PO4 and Ba(NO3)2:
3 Na3PO4 + 2 Ba(NO3)2 → Ba3(PO4)2 + 6 NaNO3
According to the balanced equation, the stoichiometric ratio between Na3PO4 and Ba(PO4)2 is 3:1. This means that for every 3 moles of Na3PO4, we can expect 1 mole of Ba(PO4)2 to be formed.
To determine the number of moles of Na3PO4, we need to divide the given mass by the molar mass of Na3PO4. The molar mass of Na3PO4 can be calculated as follows:
Na: 22.99 g/mol (per atom)
P: 30.97 g/mol (per atom)
O: 16.00 g/mol (per atom)
Molar mass of Na3PO4 = (3 x 22.99) + 30.97 + (4 x 16.00) = 163.94 g/mol
Now we can calculate the number of moles of Na3PO4:
Number of moles = Mass / Molar mass
Number of moles of Na3PO4 = 3.50 g / 163.94 g/mol
Next, we can use the stoichiometric ratio from the balanced equation to calculate the number of moles of Ba(PO4)2 that can be formed. Since the ratio is 3:1, the number of moles of Ba(PO4)2 formed will be one-third of the number of moles of Na3PO4.
Number of moles of Ba(PO4)2 = (1/3) x Number of moles of Na3PO4
Finally, to find the mass of Ba(PO4)2 formed, we multiply the number of moles of Ba(PO4)2 by its molar mass. The molar mass of Ba(PO4)2 can be calculated as:
Ba: 137.33 g/mol (per atom)
P: 30.97 g/mol (per atom)
O: 16.00 g/mol (per atom)
Molar mass of Ba(PO4)2 = 137.33 + (2 x 30.97) + (8 x 16.00) = 601.93 g/mol
Mass of Ba(PO4)2 formed = Number of moles of Ba(PO4)2 x Molar mass of Ba(PO4)2
By following these steps and plugging in the values, you can calculate the exact mass of Ba(PO4)2 that can be formed.
No Ba(PO4)2 is formed. Ba3(PO4)2 is formed.
This is a stoichiometry problem. All such problems are worked alike. Print this out and save it for future reference.
1. write the equation and balance it.
2Na3PO4 3Ba(NO3)2 ==> Ba3(PO4)2 + 6NaNO3
2. Convert 3.50 g Na3PO4 to moles. moles = grams/molar mass
3. Using the coefficients in the balanced equation, convert moles Na3PO4 to moles Ba3(PO4)2.
4. Now convert moles to grams.
g Ba3(PO4)2 = moles x molar mass.