The gram molecular mass of Cl2 is 70.906, twice the Atomic Mass of chlorine atoms. Therefore, 79.3 grams of Cl2 contains 79.3/70.906 or 1.12 moles, to the justified number of significant digits.
9.02 X 10^23 atoms Cl2 (1mol Cl2/6.022 X 10^23) = 1.50 moles Cl2
To find the number of molecules in 42.0g of Cl2, you first need to determine the number of moles using the molar mass of Cl2 (71 g/mol). Next, you can use Avogadro's number (6.022 x 10^23 molecules/mol) to convert moles to molecules. Therefore, in 42.0g of Cl2, there would be approximately 3.56 x 10^23 molecules.
The molar ratio of Cl2 to NaCl is 1:2, so for every 1 mole of Cl2, 2 moles of NaCl are produced. To find the amount of NaCl produced from 13g of Cl2, first calculate the number of moles of Cl2 using its molar mass, then use the mole ratio to determine the moles of NaCl, and finally convert to grams of NaCl.
4.005
You cannot produce any Iodine from chlorine, because chlorine (Cl2, gas) is an element, hence it does not contain any Iodine (I2, solid with purple vapor). However when 8.00 moles Cl2 react with excess (>16) moles potassium Iodide (KI) then also 8.00 moles of Iodine are produced, not FROM but BY MEANS OF chlorine. Cl2 + 2KI --> 2 KCl + I2
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.
H2 +Cl2---------------->2HCl Since H2 and Cl2 react in 1:1 mole ratio the number of moles of H2 reacting is equal to the number of moles of Cl2 which is equal to 0.213
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.
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.
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.
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.
Balanced equation: 2Al + 3Cl2 -> 2AlCl3For every 3 moles of Cl2, 2 moles of AlCl3 is produced (Using the numbers in front of the compounds)Now set up a proportion: 3/2 = 0.30/?Cross Multiply: (2 X 0.30) / 3 = 0.20.2 moles of AlCl3 will be produced.
At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 liters. Therefore, 5.60 liters of Cl2 at STP would be: 5.60 L / 22.4 L/mol = 0.25 moles of Cl2.
When 4 moles of aluminum react with an excess of chlorine gas (Cl2), 4 moles of aluminum chloride are produced because the balanced chemical equation for this reaction is: 2 Al + 3 Cl2 -> 2 AlCl3 Since the mole ratio between aluminum and aluminum chloride is 2:2, it means that for every 2 moles of aluminum, 2 moles of aluminum chloride are produced.
Cl2 + 2NaBr => Br2 + 2NaCl One mole Cl2 reacts with 2 moles NaBr Cl2 = 71 NaBr = 102.9 Molar volume = 22.414 L/mole for ideal gas @STP 3L Cl2 = 3/22.414 = 0.1338 mole 25g NaBr = 25/102.9 = 0.2430 mole 0.1338 moles Cl2 requires 0.2676 moles NaBr for complete reaction The NaBr is the limiting reagent
9.02 X 10^23 atoms Cl2 (1mol Cl2/6.022 X 10^23) = 1.50 moles Cl2
To find the number of molecules in 42.0g of Cl2, you first need to determine the number of moles using the molar mass of Cl2 (71 g/mol). Next, you can use Avogadro's number (6.022 x 10^23 molecules/mol) to convert moles to molecules. Therefore, in 42.0g of Cl2, there would be approximately 3.56 x 10^23 molecules.