the question reads:
a) balance the equation:
H3PO4 + Ca(OH)2 --> Ca3(PO4)2 + H2O
b)what mass of each product results if 750 mL of 6.00 M H3PO4 reacts according to the equation?
i balanced it with:
2H3PO4 + 3Ca(OH)2 --> Ca3(PO4)2 + 6H2O
but i'm not sure where to go from there
also,
A solution is made by dissolving 26.42 g of (NH4)2SO4 in enough H2O to make 50.00 mL of solution.
a) what is the molar mass of (NH4)2SO4?
b) what are the products of the solution?
c) how many milliliters of solution are needed?
i got:
a) 196g/mol
b) (NH4)2SO4 + H2O --> NH4O + SO4H
c) 2.7 M
i'm sure a and c is right.. just unsure about b
sorry for the long post, thanks !
H3PO4 equation is balanced.
Convert what you have to mols. M x L = mols.
Convert mols of what you have to mols of what you want.
Convert mols of what you want to grams.
grams = mols x molar mass.
For the second question, which is incomplete,
a is 132.141 but check my work since it doesn't agree with yours.
b. You haven't reacted it with anything but water, It will ionize into NH4^+ and SO4^=
c. How many mL are needed for what??
What I would first is to find the amount of mols of H3PO4. Since you know that 1 M means there is 1 mol of solution in every 1 liter, you can figure out the moles of the solution. The solution is 6M, so that means that to find the moles of solute in the solution you multiply .750 (750 milliliters converted to liters) by 6. I think you did something similar, but .0075 L makes 7.5 milliliters. This will give you 4.5 mols of H3PO4.
Now you can use stoichiometry to find the grams of Ca3(PO4)2. The balanced equation states that it takes 2 mols of H3PO4 to get 1 mol of Ca3(PO4)2:
4.5 mol H3PO4 * 1 mol Ca3(PO4)2 / 2 mol H3PO4 = 2.25 mol Ca3(PO4)2
Find the molar mass of Ca3(PO4)2 to convert the mols to grams:
3*40 + 2*31 + 8*16 = 310 grams / mol
2.25 mol Ca3(PO4)2 * 310 grams Ca3(PO4)2 / 1 mol Ca3(PO4)2 = 697.5 grams Ca3(PO4)2
You can use the same process to find the mass of water.
4.5 mol H3PO4 * 6 mol H2O / 2 mol H3PO4 = 13.5 mol H2O
13.5 mol H2O * 18 g H2O / 1 mol H2O = 243 g H2O
Hope that helped! NB not my work. its from another website
No need to apologize for the long post! Let's break it down:
For the first question:
a) You correctly balanced the equation as 2H3PO4 + 3Ca(OH)2 --> Ca3(PO4)2 + 6H2O.
b) To find the mass of each product, you need to use stoichiometry. First, calculate the number of moles of H3PO4 using the given volume and concentration:
750 mL * 6.00 mol/L = 4500 mmol = 4.50 mol
From the balanced equation, we can see that 2 moles of H3PO4 react to produce 6 moles of H2O. So, using a mole ratio:
4.50 mol H3PO4 * (6 mol H2O / 2 mol H3PO4) = 13.5 mol H2O
Now, since the molar mass of water (H2O) is approximately 18.015 g/mol, we can calculate the mass of H2O:
13.5 mol H2O * 18.015 g/mol = 243.2025 g
Similarly, using the mole ratio from the balanced equation, we can calculate the mass of Ca3(PO4)2:
4.50 mol H3PO4 * (1 mol Ca3(PO4)2 / 2 mol H3PO4) * (310.18 g/mol Ca3(PO4)2) = 668.39 g
So, the mass of each product is:
H2O: 243.20 g
Ca3(PO4)2: 668.39 g
For the second question:
a) The molar mass of (NH4)2SO4 can be calculated by adding up the atomic masses of each element:
(NH4)2SO4 = (2 * 14.01 g/mol N) + (8 * 1.01 g/mol H) + (32.07 g/mol S) + (4 * 16.00 g/mol O) = 132.14 g/mol
b) The products of the solution can be written as:
(NH4)2SO4 + H2O
c) The amount of solution needed is given as 50.00 mL, which is the volume of the final solution.
So, it seems like you had the correct answers for parts a) and c), but for part b), the products are (NH4)2SO4 and H2O.
For the balanced equation:
a) You have correctly balanced the equation as 2H3PO4 + 3Ca(OH)2 → Ca3(PO4)2 + 6H2O.
b) To find the mass of each product, we first need to determine the number of moles of H3PO4 present in the given volume (750 mL) of 6.00 M H3PO4 solution.
The number of moles can be calculated using the formula:
moles = concentration (M) × volume (L)
Since the volume is given in milliliters (mL), we need to convert it to liters (L):
750 mL = 750 / 1000 = 0.75 L
Now, calculate the number of moles of H3PO4:
moles = 6.00 M × 0.75 L = 4.50 moles
Using the balanced equation, we can determine the stoichiometric ratio between H3PO4 and the products. From the balanced equation, we see that for each 2 moles of H3PO4, we will get 1 mole of Ca3(PO4)2 and 3 moles of H2O.
So, for 4.50 moles of H3PO4, we will get:
Ca3(PO4)2: 4.50 moles × (1 mole / 2 moles) = 2.25 moles
H2O: 4.50 moles × (3 moles / 2 moles) = 6.75 moles
To find the mass of each product, we need to multiply the number of moles by their respective molar masses:
Mass of Ca3(PO4)2 = 2.25 moles × (3 × atomic mass of Ca + atomic mass of P + 4 × atomic mass of O)
Mass of H2O = 6.75 moles × (2 × atomic mass of H + atomic mass of O)
Calculate these values using the atomic masses of the elements and you will have the mass of each product.
For the second question:
a) The molar mass of (NH4)2SO4 can be calculated by summing the atomic masses of the elements present in the compound. The atomic masses (in g/mol) are as follows:
- Nitrogen (N): 14.01 g/mol
- Hydrogen (H): 1.01 g/mol
- Sulfur (S): 32.07 g/mol
- Oxygen (O): 16.00 g/mol
Molar mass of (NH4)2SO4 = 2 × (atomic mass of N) + 8 × (atomic mass of H) + atomic mass of S + 4 × (atomic mass of O)
b) When (NH4)2SO4 is dissolved in water, it ionizes to form NH4+ and SO4^2- ions. Therefore, the product of the solution is NH4+ and SO4^2- ions in the aqueous state.
c) To determine the volume of solution needed, we need the density or concentration of the solution. Given that the mass of (NH4)2SO4 is 26.42 g and the volume of the solution is 50.00 mL, we cannot directly calculate the concentration or the amount of (NH4)2SO4 dissolved. More information is needed to answer this question accurately.
Please note that in question b) for the equation H3PO4 + Ca(OH)2 --> Ca3(PO4)2 + H2O, the correct products are Ca3(PO4)2 and 4H2O, not 6H2O as mentioned in my previous answer.
No problem! Let's break down both questions:
a) Balancing the equation:
You correctly balanced the equation as:
2H3PO4 + 3Ca(OH)2 --> Ca3(PO4)2 + 6H2O
To determine the mass of each product, we need to use the concept of stoichiometry. We'll start by converting the given volume of H3PO4 solution to moles using its molarity:
Volume of H3PO4 = 750 mL = 0.750 L (since 1 L = 1000 mL)
Molarity of H3PO4 = 6.00 M
Number of moles of H3PO4 = Molarity × Volume = 6.00 mol/L × 0.750 L = 4.50 mol
Now we can use the balanced equation to determine the molar ratio between H3PO4 and the other substances:
From the balanced equation, we see that 2 moles of H3PO4 react to produce 1 mole of Ca3(PO4)2 and 6 moles of H2O.
Using this information, we can calculate the number of moles of Ca3(PO4)2 and H2O:
- Moles of Ca3(PO4)2 = (4.50 mol H3PO4) × (1 mol Ca3(PO4)2 / 2 mol H3PO4)
- Moles of H2O = (4.50 mol H3PO4) × (6 mol H2O / 2 mol H3PO4)
Now, to determine the mass of each product, we need to know their respective molar masses:
- Molar mass of Ca3(PO4)2 = 310.18 g/mol
- Molar mass of H2O = 18.02 g/mol
Finally, we can calculate the mass of each product:
- Mass of Ca3(PO4)2 = Moles of Ca3(PO4)2 × Molar mass of Ca3(PO4)2
- Mass of H2O = Moles of H2O × Molar mass of H2O
b) The correct products of the solution in question (NH4)2SO4 are:
(NH4)2SO4(aq) → 2NH4+(aq) + SO4^2-(aq)
c) In the second question, we are given the mass of (NH4)2SO4 and need to find the volume (in mL) of the resulting solution. To do this, we first need to calculate the molarity (M) of the solution:
Given: Mass of (NH4)2SO4 = 26.42 g
Molar mass of (NH4)2SO4 = 196 g/mol
Volume of solution = 50.00 mL = 0.050 L (since 1 L = 1000 mL)
Moles of (NH4)2SO4 = Mass / Molar mass
Molarity = Moles of solute / Volume of solution
Using these formulas, we can solve for the molarity of the solution:
Moles of (NH4)2SO4 = 26.42 g / 196 g/mol
Moles of (NH4)2SO4 = 0.1344 mol
Molarity = 0.1344 mol / 0.050 L
Molarity = 2.69 M (rounded to two decimal places)
So, the correct value for the molarity (in part c) is approximately 2.69 M.
I hope this helps! Let me know if you have any further questions.