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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 Cl2 molecules are in 985g of Cl2 What is the mass of Cl2 atoms present?
To find the number of moles of Cl₂ in 985 g, use the molar mass of Cl₂, which is about 70.906 g/mol. The number of moles is calculated by dividing the mass by the molar mass: [ \text{moles of Cl₂} = \frac{985 \text{ g}}{70.906 \text{ g/mol}} \approx 13.9 \text{ moles} ] Since Cl₂ consists of two chlorine atoms, the mass of Cl atoms present can be calculated by multiplying the moles of Cl₂ by 2 and then by the molar mass of chlorine (approximately 35.453 g/mol), resulting in approximately 985 g of Cl atoms.
How many moles of AlCl3 are produced when 3.64 moles of HCl reacts?
Step 1. Find limiting agent (g of H2 / amu) and (g of Cl2 / amu) Cl2 is less mole so it is limiting Step 2. Find the ratio (1:2) 1H + 1Cl = 2 HCl Step 3 mole of limiting x ratio (2) x amu of HCl Source Mastering Chemistry
How many grams of 5 moles of lead?
5 moles of lead is equal to 1 036 g.
How many moles are present in gram's of silicon Si?
1 g silicon is equal to 0,0356 moles.
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 grams cl2 needed to produce 1.5 moles of chloroform?
To produce 1.5 moles of chloroform (CHCl3), you would need 3 moles of chlorine (Cl2) as the reaction is 1:1 between Cl2 and CHCl3. The molar mass of Cl2 is approximately 70.9 g/mol, so 3 moles of Cl2 would be 3 * 70.9 g. Therefore, you would need approximately 212.7 grams of Cl2.
How many moles of cl2 are there in 7.1g of chlorine?
To find the number of moles of Cl2 in 7.1g of chlorine, you need to divide the mass of Cl2 by its molar mass. The molar mass of Cl2 is 70.9 g/mol. Therefore, 7.1g / 70.9 g/mol = 0.1 moles of Cl2.
How many grams are in NaCl with 2.34 moles of Cl2?
To find the grams of NaCl with 2.34 moles of Cl2, you need to consider the molar ratio. For every 1 mole of Cl2, there are 2 moles of Cl in NaCl. So, 2.34 moles of Cl2 would be equivalent to 4.68 moles of Cl in NaCl. Using the molar mass of NaCl (58.44 g/mol), you can calculate that 4.68 moles of NaCl would be approximately 273.64 grams.
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.
In the gaseous state chlorine exists as a diatomic molecule Cl2 (Molar mass 70.9 gmol). Calculate the number of moles of chlorine present in 140 g of chlorine gas.?
To calculate the number of moles in 140 g of Cl2, divide the given mass by the molar mass of Cl2. Number of moles = Mass / Molar mass = 140 g / 70.9 g/mol = 1.97 moles. Therefore, there are 1.97 moles of chlorine gas in 140 g of Cl2.
How many moles are in 47g of chlorine gas?
4.005
Which best represents the number of moles in exactly 84 grams of chlorine (Cl2) gas?
The molar mass of chlorine gas (Cl2) is 70.91 g/mol. To convert grams to moles, you divide the mass (84 g) by the molar mass (70.91 g/mol). So, 84 grams of chlorine gas is equal to approximately 1.18 moles.
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 limiting reactant in 2Al plus 3Cl2 equals 2AlCl3?
moles of Al=4.40 g/26.9815 g/mol=0.163 moles cl2=15.4g/70.906g/mol=0.217 the ratio is 2:3 cl2 is the limiting reagent
How many moles of Na2SO3 in 1 g?
1 g of sodium sulfite is equivalent to 0,0079 moles.
Given The Equation 2 Al 3 Cl2 gives 2 AlCl3. What mass of product in grams can be produced from 4.40 g Al and 14.4 g Cl2?
To find the limiting reactant, first convert the masses of Al and Cl2 to moles. The balanced equation shows that 2 moles of Al reacts with 3 moles of Cl2 to produce 2 moles of AlCl3. Determine the moles produced by each reactant and find the limiting reactant. Finally, calculate the mass of AlCl3 produced using the limiting reactant.
