Loading...

79.3 grams Cl2 (1mol/70.9 grams ) = 1.12 moles

319g Cl2 are needed.

Balanced equation. 2P + 5Cl2 -> 2PCl5 56.0 grams Cl2 (1 mole Cl2/70.9 grams)(2 mole PCl5/5 moles Cl2) = 0.316 moles PCl5 produced

The number of moles in exactly 84 grams of chlorine (Cl2) gas 2,37.

The mass of NaCl is 280 g.

First realize that it takes 3 CL2 to produce one mole of CHCl3. So take 1.50 moles of CHCl3 * 3 moles of Cl2 / 1 mole of CHCL3= 4.5 mole of Cl2 The the atomic weight of Cl2 is 35.5*2=71g. 4.5 mole of Cl2*71g= 319.5g. of Cl2 needed.

.807 mol

0.288 mole

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.

The molar mass of Cl2 is 5.1 g/mol. If there is 51.0 g of Cl2 then there are 10 moles. 10 moles of Cl2 can produce 4 moles of PCl5.

64 moles of chlorine atoms would be 64 x 35.5 grams, i.e. 2272 grams. If you meant the substance chlorine, which consists of Cl2 molecules, it would be 4544 grams.

12

Some conversion required.30 pounds Cl2 (454 grams/1 pound) = 13620 grams Cl213620 grams Cl2 (1 mole Cl2/70.9 grams) = 192.1 moles Cl270o F = 21.1o C = 294.3 Kelvinuse,PV = nRT(743 mm Hg)(X Volume) = (192.1 moles Cl2)(62.36 L*torr/mol*K)(294.3 K)Volume = 3525524.47/743= 4.7 X 103 Liters of chlorine gas------------------------------------------

17.7

9.02 X 10^23 atoms Cl2 (1mol Cl2/6.022 X 10^23) = 1.50 moles Cl2

Balanced equation first. Cl2 + H2 >> 2HCl Now use PV = nRT to find moles Cl2 as this limits and drives reaction. (1 atm)(5.00 L) = n(0.08206 Latm/molK)(298.15 K) = 0.204 moles Cl2 ( now use stoichiometry ) 0.204 moles Cl2 (2mole HCl/1 mole Cl2) = 0.408 moles HCl produced

1 mole of Cl2. Assuming Fe + Cl2 -> FeCl2

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 first. 2Na + Cl2 -> 2NaCl 35 grams NaCl (1 mole NaCl/58.44 grams )( 1 mole Cl2/2 mole NaCl )( 70.9 grams Cl2/1 mole Cl2) = 21 grams of Cl2 needed

2Na + Cl2 >> 2NaCl 5 moles Na (1mol Cl2/2 mole Na) = 2.5 moles Cl2 needed for this reaction.

4 moles

4 moles

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

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

0.839