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
The molarity is 2,973.
A + 2B = 3C 5A + 10B = 15C 4A + 8B = 12C A and B are the reactants. As there is not enough B to use all the A, B is the limiting factor in the production of C.
5.0
In 500ml there are 3.5 grams. Therefore in 1 litre, there are 7g. 7/58.443 is 0.1198 molar.
0.0296 m Cr
Glucose
Nitrogen is the limiting reactant and 4.15g of ammonia are produced.
I Don't knows Sorry
cuo
The first solution (500 g/L) is more concentrated than the second (200 g/L).
3,44 moles H2 react with 1,146 moles NH3. The limiting reactant is hydrogen. O,244 moles N2 remain. 19,5 g NH3 are obtained.
Ok, lets begin by writing out the reaction : 2AgNO3 +CaCl2 --> 2AgCl(s) + Ca(NO3)2 Precipitate = AgCl Now find the mol of compound in each solution: 14g AgNO3 x (mol/170g) = .082mol 4.83g CaCl2 x (mol/111g) = .044mol Determine limiting reactant: Notice in reaction that 2 CaCl2 molecules react with 1 AgNO3. Because 2(.044mol) > 1(.082mol), AgNO3 is your limiting reactant. Now that you know this you can find the mass of the precipitate .082molAgNO3x (2molAgCl/2molAgNO3)x(143.3g/molAgCl) = 11.75g b) Assuming all the AgNO3 is exhausted, there will be 2(.044)-(.082) = .006mol CaCl2 left .006mol x (111g/mol) = 0.67g CaCl2
Only a compound has a molar mass not a solution.
4.32
The molarity is 2,973.
A + 2B = 3C 5A + 10B = 15C 4A + 8B = 12C A and B are the reactants. As there is not enough B to use all the A, B is the limiting factor in the production of C.
A solution containing 5 g sodium chloride in 10 g water doesn't exist.