Calcium carbonate reacts with phosphoric acid to produce calcium phosphate, carbon dioxide, and water.
3 CaCO3(s) + 2 H3PO4(aq) Ca3(PO4)2(aq) + 3 CO2(g) + 3 H2O(l)
How many grams of phosphoric acid react with excess calcium carbonate to produce 3.29 g Ca3(PO4)2?
To calculate the grams of phosphoric acid that react with excess calcium carbonate to produce 3.29 g of Ca3(PO4)2, we need to use the stoichiometry of the reaction.
1. Start by finding the molar mass of Ca3(PO4)2:
Ca3(PO4)2:
3 Ca -> 3 * 40.08 g/mol = 120.24 g/mol
2 P -> 2 * 30.97 g/mol = 61.94 g/mol
8 O -> 8 * 16.00 g/mol = 128.00 g/mol
Total molar mass = 120.24 + 61.94 + 128.00 = 310.18 g/mol
2. Next, calculate the moles of Ca3(PO4)2 using its molar mass:
Moles = Mass / Molar mass
Moles = 3.29 g / 310.18 g/mol = 0.0106 mol
3. Now, use the stoichiometric ratio from the balanced equation to find the number of moles of phosphoric acid:
From the balanced equation, we know that the ratio of Ca3(PO4)2 to H3PO4 is 1:2.
This means that 1 mol of Ca3(PO4)2 requires 2 mol of H3PO4.
Therefore, the moles of H3PO4 will be twice the moles of Ca3(PO4)2:
Moles of H3PO4 = 2 * 0.0106 mol = 0.0212 mol
4. Finally, calculate the grams of phosphoric acid:
Grams = Moles * Molar mass
Grams = 0.0212 mol * 98.00 g/mol (molar mass of H3PO4)
Grams = 2.08 g
So, approximately 2.08 grams of phosphoric acid react with excess calcium carbonate to produce 3.29 grams of Ca3(PO4)2.
To determine the mass of phosphoric acid required to produce 3.29 g of calcium phosphate (Ca3(PO4)2), we need to use the stoichiometry of the reaction.
First, let's calculate the molar mass of calcium phosphate (Ca3(PO4)2):
- Mass of 3 calcium atoms = 3(40.08 g/mol) = 120.24 g/mol
- Mass of 2 phosphorus atoms = 2(30.97 g/mol) = 61.94 g/mol
- Mass of 8 oxygen atoms = 8(16.00 g/mol) = 128.00 g/mol
- Total molar mass of Ca3(PO4)2 = 120.24 + 61.94 + 128.00 = 310.18 g/mol
Next, let's use stoichiometry to relate the mass of calcium phosphate (Ca3(PO4)2) produced to the mass of phosphoric acid (H3PO4) reacted.
From the balanced equation, we can see that:
- 2 moles of H3PO4 react with 1 mole of Ca3(PO4)2
- The molar mass of H3PO4 is approximately 98.00 g/mol
Now, we can set up the following proportion to find the mass of phosphoric acid:
(98.00 g H3PO4 / 310.18 g Ca3(PO4)2) = (x g H3PO4 / 3.29 g Ca3(PO4)2)
Solving for x, we have:
x = (3.29 g Ca3(PO4)2 * 98.00 g H3PO4) / 310.18 g Ca3(PO4)2
Calculating this expression, we get:
x ≈ 1.04 g H3PO4
Therefore, approximately 1.04 grams of phosphoric acid would react with excess calcium carbonate to produce 3.29 grams of calcium phosphate.
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http://www.jiskha.com/display.cgi?id=1270062101