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Molar mass = 31+ 5(35.5) = 208.5 g/mole

divide ... 35.6 g / 208.5 g/mole = 0.171 mole

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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 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.


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 55.0 g of Cl2 (and excess P4)?

55.0 g of Cl2 contains 55.0/35.45* or 1.551 gram atoms of chlorine. Each mole of PCl5 requires exactly 5 gram atoms of chlorine, as shown by the formula. Therefore, 1.551/5.000 or 0.310 moles of PCl5 can be formed, to the justified number of significant digits. *This number is the gram Atomic Mass of chlorine.


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 is the decomposition reaction for PCl5?

The decomposition reaction for PCl5 is represented as follows: PCl5 (s) → PCl3 (g) + Cl2 (g). This reaction involves the breaking down of solid phosphorus pentachloride (PCl5) into gaseous phosphorus trichloride (PCl3) and chlorine gas (Cl2). The reaction is endothermic, requiring energy input to break the bonds within the PCl5 molecule.


How much energy is absorbed when 356 g of ethanol 46.0 g molboils The heat of vaporization of ethanol is 38.6 kJ mol?

To calculate the energy absorbed, first convert the mass of ethanol from grams to moles. 356 g of ethanol is 356/46.0 = 7.74 moles. Then, multiply the moles of ethanol by the heat of vaporization: 7.74 mol * 38.6 kJ/mol = 298.56 kJ of energy absorbed.


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The formula is: number of moles = g Be/9,012.


How many moles are in 978 g Ca?

978 g calcium contain 24,4 moles.


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29,0 g of calcium is equal to 0,723 moles.


How many moles are in 67.4 g of HCI?

67,4 g HCl is equivalent to 1,85 moles.