Consider the following unbalanced equation:

Ca3(Po4)2+H2SO-->CaSO4+H3PO
What masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0 kg calcium phosphate with 1.0 kg concentrated sulfuric acid (98% H2SO4 by mass)?

I'm so lost. Please help.

First you need to balance the equation AND look to make sure the subscripts are there. I assume you just made some typos when you posted. This is a limiting reagent problem; therefore, it actually is two problems in one.

1. Convert what you have to mols. mols = g/molar mass.
2. Using the coefficients in the balanced equation, convert mols of each of the starting materials to mols of the product (CaSO4). You will get two values; i.e., one form H2SO4 and one from Ca3(PO4)2. Take the smaller one and convert that to grams of the product.
3. Follow the same procedure for the H3PO4 product.Post your work if you get stuck.

Oh, don't worry! I'm here to help, and I promise not to be too "un-balanced" with my answer. Let's tackle this step by step, shall we?

First things first, let's balance that equation. It should actually be:

3Ca3(PO4)2 + 2H2SO4 → 6CaSO4 + H3PO4

Now, in order to calculate the masses of calcium sulfate (CaSO4) and phosphoric acid (H3PO4) that can be produced, we need to use stoichiometry. This involves converting the given mass of calcium phosphate (Ca3(PO4)2) and concentrated sulfuric acid (H2SO4) into moles, and then using the balanced equation to determine the moles of the desired products.

Let's start with calcium phosphate (Ca3(PO4)2):

1. Calculate the moles of calcium phosphate:

Molar mass of Ca3(PO4)2 = 310.18 g/mol
Moles = mass (g) / molar mass (g/mol)

Given that the mass of calcium phosphate is 1.0 kg = 1000 g,
Moles of Ca3(PO4)2 = 1000 g / 310.18 g/mol

2. Now, let's consider sulfuric acid (H2SO4):

The molar mass of sulfuric acid (H2SO4) = 98.09 g/mol.

Since sulfuric acid concentration is given as 98% by mass, we can calculate the mass of pure sulfuric acid (H2SO4) present in the 1.0 kg solution:

Mass of H2SO4 = 1.0 kg × 0.98 = 980 g

Again, let's convert this mass into moles:
Moles of H2SO4 = 980 g / 98.09 g/mol

3. Next, we use the balanced equation to determine the mole ratio between calcium phosphate and the desired products (calcium sulfate and phosphoric acid). From the balanced equation, we see that:

3 moles Ca3(PO4)2 yield 6 moles CaSO4
3 moles Ca3(PO4)2 yield 1 mole H3PO4

So, using the moles of Ca3(PO4)2 calculated earlier, we can determine the moles of the desired products:

Moles of CaSO4 = Moles of Ca3(PO4)2 × (6 moles CaSO4 / 3 moles Ca3(PO4)2)
Moles of H3PO4 = Moles of Ca3(PO4)2 × (1 mole H3PO4 / 3 moles Ca3(PO4)2)

4. Finally, we convert the moles of each product back into mass:

Mass of CaSO4 = Moles of CaSO4 × molar mass of CaSO4
Mass of H3PO4 = Moles of H3PO4 × molar mass of H3PO4

And voila! You'll have the masses of calcium sulfate and phosphoric acid that could be produced from the given reaction.

I hope that helps, and remember, chemistry may have its "un-balanced" moments, but with a little patience, we can figure it out!

To solve this problem, we first need to balance the given chemical equation:

Ca3(PO4)2 + H2SO4 → CaSO4 + H3PO4

Now, let's calculate the molar masses of each compound involved:

Ca3(PO4)2: (3*40.08) + (2*31.00) + (8*16.00) = 310.18 g/mol
H2SO4: (2*1.01) + 32.07 + (4*16.00) = 98.09 g/mol
CaSO4: 40.08 + 32.07 + (4*16.00) = 136.14 g/mol
H3PO4: (3*1.01) + 30.97 + (4*16.00) = 97.99 g/mol

Now, let's determine the limiting reagent by comparing the number of moles of calcium phosphate (Ca3(PO4)2) and sulfuric acid (H2SO4) in 1.0 kg of each compound:

Mass of Ca3(PO4)2 = 1.0 kg = 1000 g
Number of moles of Ca3(PO4)2 = 1000 g / 310.18 g/mol ≈ 3.22 mol

Since the 1.0 kg concentrated sulfuric acid is 98% H2SO4 by mass, we need to calculate the mass of H2SO4 present in 1.0 kg of the acid:

Mass of H2SO4 = 0.98 * 1.0 kg = 0.98 kg = 980 g
Number of moles of H2SO4 = 980 g / 98.09 g/mol ≈ 9.99 mol

From the equation, we can see that the ratio of Ca3(PO4)2 to H2SO4 is 1:1. Therefore, the limiting reagent is calcium phosphate (Ca3(PO4)2) because it has fewer moles.

To determine the mass of calcium sulfate (CaSO4) produced, we use the stoichiometry of the balanced equation:

1 mol Ca3(PO4)2 produces 1 mol CaSO4
3.22 mol Ca3(PO4)2 produces (3.22 mol CaSO4) / (1 mol Ca3(PO4)2) = 3.22 mol CaSO4

Mass of CaSO4 = (molar mass of CaSO4) * (number of moles of CaSO4)
= 136.14 g/mol * 3.22 mol
≈ 438.63 g

To determine the mass of phosphoric acid (H3PO4) produced, we use the stoichiometry of the balanced equation:

1 mol Ca3(PO4)2 produces 1 mol H3PO4
3.22 mol Ca3(PO4)2 produces (3.22 mol H3PO4) / (1 mol Ca3(PO4)2) = 3.22 mol H3PO4

Mass of H3PO4 = (molar mass of H3PO4) * (number of moles of H3PO4)
= 97.99 g/mol * 3.22 mol
≈ 315.35 g

Therefore, from the reaction of 1.0 kg of calcium phosphate with 1.0 kg of concentrated sulfuric acid, approximately 438.63 g of calcium sulfate and 315.35 g of phosphoric acid can be produced.

To solve this problem, we need to balance the given chemical equation first.

The unbalanced equation is:
Ca3(PO4)2 + H2SO4 → CaSO4 + H3PO4

Balancing the equation:
3 Ca3(PO4)2 + 2 H2SO4 → 3 CaSO4 + 2 H3PO4

Now, let's calculate the molar masses of the compounds involved to determine the number of moles.

Molar mass of Ca3(PO4)2:
3(40.08 g/mol) + 2(31.00 g/mol) + 8(16.00 g/mol) + 2(4(1.01 g/mol) + 16.00 g/mol) = 310.18 g/mol

Molar mass of H2SO4:
2(1.01 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 98.09 g/mol

Now we can calculate the number of moles for each compound:

Number of moles of Ca3(PO4)2:
1.0 kg / 310.18 g/mol = 3.219 mol

Number of moles of H2SO4:
1.0 kg x 0.98 (as the concentrated sulfuric acid is 98% H2SO4 by mass) / 98.09 g/mol = 9.999 mol (approximately 10 mol)

According to the balanced equation, the stoichiometric ratio between Ca3(PO4)2 and CaSO4 is 3:3, which means 3 moles of Ca3(PO4)2 will produce 3 moles of CaSO4.

So, from the 3.219 moles of Ca3(PO4)2, we can expect to get 3.219 moles of CaSO4.

Number of moles of CaSO4 produced = 3.219 mol

The molar mass of CaSO4 is:
1(40.08 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 136.14 g/mol

Now, let's calculate the mass of CaSO4 produced:
Mass of CaSO4 = Number of moles of CaSO4 × Molar mass of CaSO4
= 3.219 mol × 136.14 g/mol
= 438.58 g

So, from the reaction of 1.0 kg of calcium phosphate, we can obtain approximately 438.58 g of calcium sulfate (CaSO4).

Similarly, let's calculate the mass of phosphoric acid (H3PO4) produced.

According to the balanced equation, the stoichiometric ratio between Ca3(PO4)2 and H3PO4 is 2:2, which means 2 moles of Ca3(PO4)2 will produce 2 moles of H3PO4.

So, from the 3.219 moles of Ca3(PO4)2, we can expect to get 3.219 moles of H3PO4.

Number of moles of H3PO4 produced = 3.219 mol

The molar mass of H3PO4 is:
3(1.01 g/mol) + 31.00 g/mol + 4(16.00 g/mol) = 98.00 g/mol

Now, let's calculate the mass of H3PO4 produced:
Mass of H3PO4 = Number of moles of H3PO4 × Molar mass of H3PO4
= 3.219 mol × 98.00 g/mol
= 315.76 g

So, from the reaction of 1.0 kg of calcium phosphate, we can obtain approximately 315.76 g of phosphoric acid (H3PO4).

To summarize, when 1.0 kg of calcium phosphate reacts with 1.0 kg of concentrated sulfuric acid (98% H2SO4 by mass), approximately 438.58 g of calcium sulfate (CaSO4) and approximately 315.76 g of phosphoric acid (H3PO4) can be produced.