Answer:
In order to prepare 200.0 mL of an aqueous solution of iron (III) chloride, at a concentration of 1.25 x 10⁻² M, you need to weight 0.4055 g of FeCl₃ and add to 200.0 mL of water.
Explanation:
Concentration: 1.25 x 10⁻² M
1,25 x 10⁻² mol FeCl₃ ___ 1000 mL
x ___ 200.0 mL
x = 2.5 x 10⁻³ mol FeCl₃
Mass of FeCl₃:
1 mol FeCl₃ _____________ 162.2 g
2.5 x 10⁻³ mol FeCl₃ _______ y
y = 0.4055 g FeCl₃
1) first, we have to convert the grams to moles of AuCl3 using the molar mass of the molecule.
molar mass of AuCl3= 197 + (35.5 x 3)= 303.5 g/mol
73.4 g (1 mol AuCl3/ 303.5 g)= 0.242 moles
2) now. let's convert moles of AuCl3 to moles of chlorine gas (Cl2) using the mole-mole ratio
0.242 mol AuCl3 (3 mol Cl2/ 2 mol AuCl3)= 0.363 mol Cl2
3) finally, we convert moles to grams using the molar mass of Cl2.
molar mass of Cl2 = 35.5 x 2= 71.0 g/mol
0.363 mol Cl2 ( 71.0 g/ 1 mol)= 25.8 grams
Michael uses a car that burns
fossil fuels. COMBUSTION REACTION
Alan grows tomatoes on
his farm. PHOTOSYNTHESIS
Lisa eats eggs for breakfast. RESPIRATION
The balanced equation between NaOH and H₂SO₄ is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol
Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol
number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol
molarity of H₂SO₄ is 0.03617 M
