# of Moles = Mass in grams divided by Molar Mass =5o divided by (cl x 2) =50 divided by 71 =0.704 moles
use:
1 mol = Mr in grams
that is 35.5x2 g of Cl2 = 1 mol
71g of Cl2 = 1 mol
therefore 50g of Cl2 = (1/71) x 50
=0.704 mol
For this you need the atomic (molecular) mass of CaCl2. Take the number of moles and multiply it by the Atomic Mass. Divide by one mole for units to cancel. CaCl2= 111.1
.500 moles CaCl2
For this you need the atomic (molecular) mass of CaCl2. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel. CaCl2= 111.1
.500 moles CaCl2 × (111.1 grams) =55.6 grams CaCl2
There are 110.978 grams in one mole of CaCl2
Take the atomic weight of Cl, which is 35.453 g/mol. Multiply 35.453 by 2 because you have 2 Cl atoms (Cl2). You will get 70.906 g/mol. Then 1/70.906 equals 1.41x10-2. That is your answer.
50 g potassium chloride is equivalent to 0,67 moles.
50g/36.5g=1.3699 moles
.213 g Cl/35.5 g Cl X 1 mol = 6.0 moles
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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
4 g of He = 1 mole. So, 30.8 g = 7.7 moles
5 moles of lead is equal to 1 036 g.
1 g silicon is equal to 0,0356 moles.
At 22.4 liters a mole at STP, and the molar mass of Cl2 being 71, 17.32 g is about .242 moles. Multiply the moles by standard volume and you get 5.43 liters.
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
4.005
The mass of NaCl is 280 g.
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.
I2 + 2Cl2 .==> ICl + ICl3 molar mass of I2 = 127 molar mass of Cl2 = 70.9 moles of I2 used = 25.4/127 = 0.2 moles of I2 moles of Cl2 used = 14.2/70.9 = 0.2 moles of Cl2 Since the stoichiometry indicates that each mole of I2 reacts with 2 moles of Cl2 to give 1 mole each of ICl and ICl3, you should get 0.1 moles of each product. The reason you don't get 0.2 moles is because the amount of I2 is limiting, and you can see that 2 moles of Cl2 will use up 1 mole of I2, so 0.2 moles of Cl2 will use up 0.1 moles of I2. So 0.1 moles of product is all you can get under these conditions
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
1 g of gold is equal to 0,006 moles.
1 g of sodium sulfite is equivalent to 0,0079 moles.
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
2LiBr(aq) + Cl2(g) = 2LiCl(aq) + Br2(l) will result in .167 moles of lithium chloride.
H2 (g) + Cl2 (g) --> 2 HCl (g) 25.00 g HCl x 1 mol HCl x 1 mol Cl2 x 70.90 g Cl2 = 24.3 g Cl2 are needed. ................... 36.46 g HCl . 2 mol HCl .. 1 mol Cl2
Remember moles = nass(g) / Mr ( Relative Moleculr Mass) . mass = 120 g Mr =35.5 x 2 = 71 ( Because chlorine exists as a diatomic gas Cl2). Hence moles(Cl2) = 120 /71 = 1.690