Answer:
6.24%
Explanation:
Molality by definition means a measurement of the number of moles of solute in solution with 1000 gm or 1Kg solvent. Notice the difference that Molarity is defined on the volume of solution and Molality on the mass of solvent.
So, An aqueous solution of iron(II) iodide has a concentration of 0.215 molal.
means 0.215 moles are present in 1 Kg of solvent.
The molar mass of Fe2I = 309.65 g / mole
mass of FeI2 = moles x molar mass
= 0.215 x 309.65
=66.57 gm
mass % of FeI2 = mass of FeI2 x 100 / total mass
= 66.57x 100 / (1000 +66.57)
= 6.24%
Answer: The new concentration of a solution of 
is 0.2 M 10.0 mL of a 2.0 M solution is diluted to 100 mL.
Explanation:
Given: 
= 10.0 mL, 
= 2.0 M
= 100 mL, 
= ?
Formula used to calculate the new concentration is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that the new concentration of a solution of is 0.2 M 10.0 mL of a 2.0 M solution is diluted to 100 mL.
Answer:
Oxygen = 15.999 g/mol
Iron = 3 × 55.845 = 167.535 g/mol
CaCO3 = 20 × 100.0869 = 2001.738 g/mol
Answer:
2Fe + 6HC2H3O2 → 2Fe(C2H3O2)3 + 3H2
Explanation:
There you go
Answer:
Ammonia is limiting reactant
Amount of oxygen left = 0.035 mol
Explanation:
Masa of ammonia = 2.00 g
Mass of oxygen = 4.00 g
Which is limiting reactant = ?
Balance chemical equation:
4NH₃ + 3O₂ → 2N₂ + 6H₂O
Number of moles of ammonia:
Number of moles = mass/molar mass
Number of moles = 2.00 g/ 17 g/mol
Number of moles = 0.12 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 4.00 g/ 32 g/mol
Number of moles = 0.125 mol
Now we will compare the moles of ammonia and oxygen with water and nitrogen.
NH₃ : N₂
4 : 2
0.12 : 2/4×0.12 = 0.06
NH₃ : H₂O
4 : 6
0.12 : 6/4×0.12 = 0.18
O₂ : N₂
3 : 2
0.125 : 2/3×0.125 = 0.08
O₂ : H₂O
3 : 6
0.125 : 6/3×0.125 = 0.25
The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.
Amount of oxygen left:
NH₃ : O₂
4 : 3
0.12 : 3/4×0.12= 0.09
Amount of oxygen react = 0.09 mol
Amount of oxygen left = 0.125 - 0.09 = 0.035 mol
