Answer:
1.714 M
Explanation:
We'll begin by calculating the number of mole in 46.8 g of NaHCO₃. This can be obtained as follow:
Mass of NaHCO₃ = 46.8 g
Molar mass of NaHCO₃ = 23 + 1 + 12 + (3×16)
= 23 + 1 + 12 + 48
= 84 g/mol
Mole of NaHCO₃ =?
Mole = mass / molar mass
Mole of NaHCO₃ = 46.8 / 84
Mole of NaHCO₃ = 0.557 mole
Next, we shall convert 325 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
325 mL = 325 mL × 1 L / 1000 mL
325 mL = 0.325 L
Thus, 325 mL is equivalent to 0.325 L.
Finally, we shall determine the molarity of the solution. This can be obtained as shown below:
Mole of NaHCO₃ = 0.557 mole
Volume = 0.325 L
Molarity =?
Molarity = mole / Volume
Molarity = 0.557 / 0.325
Molarity = 1.714 M
Therefore the molarity of the solution is 1.714 M
The reaction between phosphoric acid and ammonia that produces ammonium phosphate can be written as follows:
3NH3 + H3PO4 ..................> (NH4)3PO4
From the periodic table:
molar mass of nitrogen = 14 grams
molar mass of hydrogen = 1 grams
molar mass of oxygen = 16 grams
molar mass of phosphorus = 30.9 grams
based on this:
molar mass of 3NH3 = 3 (14 + 3(1)) = 51 grams
molar mass of H3PO4 = 3(1) + 30.9 + 4(16) = 97.9 grams
molar mass of (NH4)3PO4 = 3 (14 + 4(1)) + 30.9 + 4(16) = 54 + 30.9 + 64
= 148.9 grams
Therefore, 97.9 grams of phosphoric acid is required to produced 148.9 grams of ammonium phosphate.
Thus, to know the mass of ammonium phosphate produced from 4.9 grams of phosphoric acid, we will simply use cross multiplication as follows:
amount of produced ammonium phosphate = (4.9 x 148.9) / 97.9 = 7.45 g
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