How many grams of calcium hydroxide will be needed to react completely with 10.0 g of phosphorus acid?
Well, let me start by saying that calcium hydroxide and phosphorus acid went on a date once. It didn't really work out because they had zero chemistry. But let's calculate anyway!
To figure out how many grams of calcium hydroxide you'll need, we need to compare the molar ratios between calcium hydroxide and phosphorus acid. The molar ratio between calcium hydroxide and phosphorus acid is 2:1.
So, we'll convert 10.0 g of phosphorus acid into moles by using its molar mass, which is approximately 98 g/mol. Dividing 10.0 g by 98 g/mol gives us around 0.102 moles of phosphorus acid.
Since the molar ratio is 2:1, you'll need half as many moles of calcium hydroxide. That means you would need approximately 0.051 moles of calcium hydroxide.
To find the grams of calcium hydroxide, we multiply the moles of calcium hydroxide by its molar mass, which is about 74 g/mol. Multiplying 0.051 moles by 74 g/mol gives us around 3.774 g of calcium hydroxide.
So, you'll need approximately 3.774 grams of calcium hydroxide to react completely with 10.0 grams of phosphorus acid. But remember, chemistry can be surprising, so be prepared for some unexpected reactions!
To determine the amount of calcium hydroxide needed to react completely with 10.0 g of phosphorus acid, we need to balance the chemical equation between calcium hydroxide (Ca(OH)2) and phosphoric acid (H3PO4). The balanced equation is:
3Ca(OH)2 + 2H3PO4 -> Ca3(PO4)2 + 6H2O
From the equation, we can see that 3 moles of calcium hydroxide react with 2 moles of phosphoric acid. To calculate the amount of calcium hydroxide needed, we'll follow these steps:
Step 1: Calculate the molar mass of phosphoric acid.
H = 1.0079 g/mol * 3 = 3.0237 g/mol
P = 30.9738 g/mol * 1 = 30.9738 g/mol
O = 15.9994 g/mol * 4 = 63.9976 g/mol
Molar mass of H3PO4 = 3.0237 + 30.9738 + 63.9976 = 97.9965 g/mol
Step 2: Calculate the number of moles of phosphoric acid.
Number of moles = mass / molar mass
Number of moles = 10.0 g / 97.9965 g/mol = 0.1021 mol
Step 3: Use the stoichiometric ratio to find the number of moles of calcium hydroxide needed.
From the balanced equation, we know that 2 moles of H3PO4 react with 3 moles of Ca(OH)2.
Therefore, the ratio is 2:3.
Number of moles of Ca(OH)2 = (3/2) * number of moles of H3PO4
Number of moles of Ca(OH)2 = (3/2) * 0.1021 mol = 0.1532 mol
Step 4: Calculate the mass of calcium hydroxide needed.
Mass of Ca(OH)2 = molar mass * number of moles
Mass of Ca(OH)2 = 74.093 g/mol * 0.1532 mol = 11.350 g
Therefore, 11.350 grams of calcium hydroxide will be needed to react completely with 10.0 grams of phosphoric acid.
To determine how many grams of calcium hydroxide will be needed to react completely with 10.0 g of phosphoric acid, we first need to write and balance the balanced chemical equation for the reaction.
The balanced equation for the reaction between calcium hydroxide (Ca(OH)2) and phosphoric acid (H3PO4) can be written as:
3 Ca(OH)2 + 2 H3PO4 → Ca3(PO4)2 + 6 H2O
From the balanced equation, we can see that 3 moles of calcium hydroxide react with 2 moles of phosphoric acid.
Now, we need to calculate the number of moles of phosphoric acid present in 10.0 g. To do this, we use the formula:
moles = mass / molar mass
The molar mass of phosphoric acid (H3PO4) can be calculated as follows:
Molar mass of H3PO4 = (3 * atomic mass of Hydrogen) + (1 * atomic mass of Phosphorous) + (4 * atomic mass of Oxygen)
= (3 * 1.01 g/mol) + (1 * 31.0 g/mol) + (4 * 16.0 g/mol)
= 98.0 g/mol
Using the formula, we can calculate the moles of phosphoric acid:
moles of H3PO4 = 10.0 g / 98.0 g/mol
= 0.102 moles
From the balanced equation, we know that 2 moles of phosphoric acid react with 3 moles of calcium hydroxide. Therefore, we can set up a proportion to find out how many moles of calcium hydroxide are required:
2 moles of H3PO4 / 3 moles of Ca(OH)2 = 0.102 moles of H3PO4 / x
Solving for x, we find:
x = (3 moles of Ca(OH)2 / 2 moles of H3PO4) * 0.102 moles of H3PO4
= 0.153 moles of Ca(OH)2
Finally, to calculate the grams of calcium hydroxide needed, we use the formula:
mass = moles * molar mass
The molar mass of calcium hydroxide (Ca(OH)2) can be calculated as:
Molar mass of Ca(OH)2 = (1 * atomic mass of Calcium) + (2 * atomic mass of Oxygen) + (2 * atomic mass of Hydrogen)
= (1 * 40.1 g/mol) + (2 * 16.0 g/mol) + (2 * 1.01 g/mol)
= 74.1 g/mol
Now, plugging in the values, we can find the grams of calcium hydroxide needed:
mass = 0.153 moles * 74.1 g/mol
= 11.3 grams
Therefore, 11.3 grams of calcium hydroxide will be needed to react completely with 10.0 grams of phosphoric acid.
I assume you meant phosphorous acid, H3PO3.
2H3PO3 + 3Ca(OH)2 ==> Ca3(PO3)2 + 6H2O
10.0 g H3PO3/82 = 0.12 mols H3PO3. Now look at the balanced equation. Convert 0.12 mols H3PO3 to mols Ca(OH)2 this way.
0.12 mols H3PO3 x (2 mols Ca(OH)2/3 mols H3PO3) = 0.08 mols Ca(OH)2 needed. Then grams Ca(OH)2 = mols x molar mass = ?
Post your work if you get stuck.