The answer is 0,3422 grams.
Given the balanced equation2Al + 6HBr --> 2AlBr3 + 3H2In order to find how many grams of HBr are required to produce 150g AlBr3, we must convert from mass to mass (mass --> mass conversion).150g AlBr3 * 1 mol AlBr3 * 6 molecules HBr = 136.52 or 137g HBr----------- 266.6g AlBr3 * 2 molecules AlBr3
SrCl2 Strontium (Sr) = 87.62 grams/mol 2 Chlorine (Cl) = 70.9 grams/mol ----------------------------------------------------add Strontium chloride = 158.52 grams/mol ============================
92.4 grams
1 Mol
This depends on the chemical compound.
3.8 g Fe * 1 mol Fe/55.85 g Fe (molar mass) = .0680 mol Fe .0680 mol Fe * 2 mol HBr/1 mol Fe (found in formula Fe+2HBr=>FeBr2+H2)=.136 mol HBr .136 mol HBr*80.912 g HBr/1 mol HBr=11.004 g HBr (or 11 using 2 sig figs) And the mass of H2 that is produced is 0.14 g
2.01 mol
1.25 mol HBr is (maximally) produced from 1.25 mol Br (better: 0.625 mol Br2) Since its molecular mass is 80.91 g mol−1 this 1.25 mol corresponds with 101.1 gram HBr
Given the balanced equation2Al + 6HBr --> 2AlBr3 + 3H2In order to find how many grams of HBr are required to produce 150g AlBr3, we must convert from mass to mass (mass --> mass conversion).150g AlBr3 * 1 mol AlBr3 * 6 molecules HBr = 136.52 or 137g HBr----------- 266.6g AlBr3 * 2 molecules AlBr3
SrCl2 Strontium (Sr) = 87.62 grams/mol 2 Chlorine (Cl) = 70.9 grams/mol ----------------------------------------------------add Strontium chloride = 158.52 grams/mol ============================
92.4 grams
769.0 grams
357
As the molar mass of NaCl is 58.5 g/mol, 1.6 moles weight is 93.6 grams.
1 Mol
42
This depends on the chemical compound.