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How Many Moles Of PCl5 Can Be Produced From 58.0 G Of Cl2 (and Excess P4)?
To determine how many moles of PCl5 can be produced from 58.0 g of Cl2, we first need to calculate the moles of Cl2. The molar mass of Cl2 is approximately 70.9 g/mol, so the number of moles of Cl2 is 58.0 g / 70.9 g/mol ≈ 0.819 moles. The balanced reaction for the formation of PCl5 from P4 and Cl2 is: P4 + 10 Cl2 → 4 PCl5. From this, we see that 10 moles of Cl2 produce 4 moles of PCl5, so 0.819 moles of Cl2 can produce (0.819 moles Cl2) * (4 moles PCl5 / 10 moles Cl2) ≈ 0.3276 moles of PCl5. Thus, approximately 0.328 moles of PCl5 can be produced.
How much energy is required to decompose 765 g of PCl3 in 4PCl3---P4 plus 6Cl2?
The balanced chemical equation is 4PCl3 → P4 + 6Cl2. The molar mass of PCl3 is 137.33 g/mol. To calculate the energy required, first find the moles of PCl3 (765 g / 137.33 g/mol), then use the stoichiometry from the balanced equation to convert moles of PCl3 to moles of P4, and finally use the enthalpy values for each reaction to determine the total energy required. The enthalpy values for each reaction can then be multiplied by the number of moles of each substance in the reaction to find the total energy required for the reaction.
P4 plus 6F2 yields 4PF3 What mass of F2 is needed to produce 186 g of PF3 if the reaction has a 76.5 percent yield?
First, calculate the molar mass of PF3: P = 30.97 g/mol, F = 19.00 g/mol, so PF3 = 30.97 + (3 * 19.00) = 88.97 g/mol. Given that the reaction is 76.5% yield, the actual mass of PF3 produced is 186 g / 0.765 = 242.75 g. From the balanced equation, 6 moles of F2 are needed to produce 4 moles of PF3. Therefore, the molar ratio is 6:4. Calculate the moles of PF3 produced and then find the moles of F2 needed. Finally, convert moles of F2 to grams.
Examples of polyatomic molecules?
Water (H2O), carbon dioxide (CO2), sulfur dioxide (SO2), ammonia (NH3), and methane (CH4) are all examples of polyatomic molecules. These molecules contain more than two atoms chemically bonded together to form a stable structure.
How many atoms are there in the molecule of phospherous P4?
The answer is four. The molecular formula of any element or compund gives you the number of atoms of each element in a molecule. For phosphorus the most common form is P4 . The molecules are tetrahedral with a P atom at each apex.
How many moles of PCl5 can be produced from 24 grams of P4?
Assuming that you are combining the P4 with Cl2 and there is a suffiecient quantity of Cl2 for the P4 to completely react, you will first need a balanced equation which is P4 + 10Cl2 -> 4PCl5. From there, it's mostly stoichiometry. Take the 24g of P4, divide by the molar mass (123.88g/mol) to get the number of moles of P4 that you have (0.194). You then have to convert, using the balanced equation, from moles of P4 to moles of PCl5, in this case multiplying by 4. That will give you the number of moles of PCl5. The stoichiometry should look something like this 24.0 g P4 x (1 mol P4/123.88g P4) x (4 mol PCl5/1 mol P4).
What mass of PCl 5 will be produced from the given masses of both reactants?
What mass of will be produced from the given masses of both reactants? 28.0 g/P4 / 123.9 g/P4 = 0.2260 moles/ P4 0.2260 moles /P4 * 4 moles PCl5/1 mole/P4=0.904 moles/PCl5 54.0 g/Cl2 / 70.9 g/Cl2 = .7616 moles/Cl2 .7616 moles / Cl2 *4 moles PCl5/10 moles Cl2=.30 This is about limiting reagents You need to use 2P + 5Cl2 --> 2PCl5 [or P4 + 10Cl2 --> 4PCl5 if you prefer] that tells you that 2P = 62g needs 5Cl2 = 355g Cl2 to react 69.3g
What mass of PCl5 will be produced from the given masses of both reactants which is 839 moles of P4 and 316 moles of Cl2 Answer needs to be submitted in grams though please?
Our lower number of moles is .316 so we will use that. The total molar mass of PCl5 is 208.2. So we will multiply the lower number of moles by our total mass. .316mol X 208.2 g/mol = 65.8g <------ answer.
How many moles of PCl5 can be produced from 56.0g Cl2 and excess P4?
To find out how many moles of PCl5 can be formed from the reaction of P4 and Cl2, it is necessary to set up the stoichiometric equation. X P4 + Y Cl2 --> Z PCl5. Balancing the equation, X = 1, Y = 10, and Z = 4. This means that for every mole of P4 that reacts, 4 moles of PCl5 is produced. The next step is to find out how many moles of P4 are present in 30.0 grams. The molar mass of P4 is 123.895 g/mol, so there are .24214 moles of P4 present. Multiplied by 4, the answer is 0.96856 moles of PCl5 are produced.
How Many Moles Of PCl5 Can Be Produced From 58.0 G Of Cl2 (and Excess P4)?
To determine how many moles of PCl5 can be produced from 58.0 g of Cl2, we first need to calculate the moles of Cl2. The molar mass of Cl2 is approximately 70.9 g/mol, so the number of moles of Cl2 is 58.0 g / 70.9 g/mol ≈ 0.819 moles. The balanced reaction for the formation of PCl5 from P4 and Cl2 is: P4 + 10 Cl2 → 4 PCl5. From this, we see that 10 moles of Cl2 produce 4 moles of PCl5, so 0.819 moles of Cl2 can produce (0.819 moles Cl2) * (4 moles PCl5 / 10 moles Cl2) ≈ 0.3276 moles of PCl5. Thus, approximately 0.328 moles of PCl5 can be produced.
What is the maximum amount in moles of P2O5 that can theoretically be made from 211 g of P 4 and excess oxygen Express your answer to three significant figures and include the appropriate units?
The molar mass of P4 is 123.9 g/mol, and the molar mass of P2O5 is 141.94 g/mol. The stoichiometry of the reaction shows that 1 mol of P4 produces 1 mol of P2O5. Therefore, the maximum amount of P2O5 that can be produced from 211 g of P4 is 211 g / 123.9 g/mol = 1.70 mol of P2O5.
What is the maximum amount of P4 that can be produced from 1.0kg of phosphorite if the phorphorite sample is 75 Ca3PO42 by mass?
2Ca3(PO4)2 + 6SiO2 + 10 C -> 6CaSiO3 + P4 + 10 CO There is 1 kg of solution. 75% of the solution is Ca3(PO4)2. Change 1kg to grams which is 1000g. 75% of 1000 is 750 Therefore there is 750g of the Ca3(PO4)2. Find the mole of Ca3(PO4)2 which is the mass divided by the molar mass. 750g/310.4gmol-1 which equals n =2.4mol ...1/2 is the ratio so mol of P4 is 1.2 mol. to find the mass of P4 you do M * n = 1.2mol * 148.65 therefore the max amount of P4 that can be produced is 150g
Phosphorus why some place denoted as P4?
Phosphorus forms individual P4 molecules. This is the standard form for white phosphorus.
A molecule of a compound?
Most molecules are compounds but not all. Some molecules such as O2 and P4 are elements.
How much energy is required to decompose 765 g of PCl3 in 4PCl3---P4 plus 6Cl2?
The balanced chemical equation is 4PCl3 → P4 + 6Cl2. The molar mass of PCl3 is 137.33 g/mol. To calculate the energy required, first find the moles of PCl3 (765 g / 137.33 g/mol), then use the stoichiometry from the balanced equation to convert moles of PCl3 to moles of P4, and finally use the enthalpy values for each reaction to determine the total energy required. The enthalpy values for each reaction can then be multiplied by the number of moles of each substance in the reaction to find the total energy required for the reaction.
How many grams of potassium cyanide PCl3 is produced from 93.0 grams of P4 (s) and 213 g of Cl2 (g) assuming the reaction goes to completion?
The balanced chemical equation for the reaction is: P4 (s) + 6Cl2 (g) -> 4PCl3 (s). To find the limiting reactant, we need to calculate the moles of P4 and Cl2 using their molar masses first. Then, we compare the amounts of PCl3 that can be produced from each reactant, and the smaller value (in moles) will be the limiting reactant. Finally, convert the moles of PCl3 to grams.
P4 plus 6F2 yields 4PF3 What mass of F2 is needed to produce 186 g of PF3 if the reaction has a 76.5 percent yield?
First, calculate the molar mass of PF3: P = 30.97 g/mol, F = 19.00 g/mol, so PF3 = 30.97 + (3 * 19.00) = 88.97 g/mol. Given that the reaction is 76.5% yield, the actual mass of PF3 produced is 186 g / 0.765 = 242.75 g. From the balanced equation, 6 moles of F2 are needed to produce 4 moles of PF3. Therefore, the molar ratio is 6:4. Calculate the moles of PF3 produced and then find the moles of F2 needed. Finally, convert moles of F2 to grams.
