The moles of oxygen gas (O2) that is needed is 4 moles
Explanation
2H2 +O2 → 2H2O
The moles of O2 is determined using the mole ratio of H2:O2
that is from equation above H2:O2 is 2:1
If the moles of H2 is 8 moles therefore the moles of O2
= 8 moles x 1/2 = 4 moles
Answer:
See explaination
Explanation:
Going by the clues that it is between Silver Flouride (AgF) and Sodium Fluoride (NaF) and since it is an aqueous solution , the 1 liter bottle is likely to be Sodium Chloride( NaCl). Going by the reaction,
AgF + NaCl= AgCl + NaF
Here, the color of AgCl is white, hence the solution cannot be AgCl.
Determination of NaCl
Determination of NaCl can be done by Mohr's Method or Volhard's method. But results in Volhard's method are more accurate . Its uses the method of back titration with Potassium Thiocynate which forms a AgCl precipitate . Prior to titration,excess AgNO3 ( The problem also has a clue that excess reagents are present in the lab ) is added to the NaCl solution so that all the Cl- ions react with Ag+. Fe3+ is then added as an indicator and the solution is titrated with KSCN to form a silver thiocyannite precipitate (AgSCN). Once all the silver has reacted, a slight excess of SCN- reacts with Fe3+ to form Fe(SCN)3 dark red complex. The concentration of Cl- is determined by subtracting the titer findings of Ag+ ions that reacted to form AgSCN from the Ag NO3 moles added to the solution. This is used because pH of the solution is acidic. If the pH of solution is basic, Mohr's method is used.
Reactions
Ag+ (aq)+ Cl-(aq) = AgCl(aq)
Ag+(aq) + SCN-(aq) = AgSCN(aq)
Fe3+(aq) + SCN-(aq) = [FeSCN]2- (aq)
Answer:
10.5 g
Explanation:
Step 1: Given data
- Molar concentration of the solution (C): 0.243 M
- Volume of solution (V): 0.580 L
Step 2: Calculate the moles of solute (n)
Molarity is equal to the moles of solute divided by the liters of solution.
M = n/V
n = M × V
n = 0.243 mol/L × 0.580 L = 0.141 mol
Step 3: Calculate the mass corresponding to 0.141 moles of KCl
The molar mass of KCl is 74.55 g/mol.
0.141 mol × 74.55 g/mol = 10.5 g
Since
potassium and phosphate is what we are to find for and they are both found in
the potassium phosphate solution, therefore we solve for this one first on the
basis of the phosphate.
The formula
for finding the volume given the concentration and number of moles is:
Volume =
number of moles / concentration in Molarity
Volume
potassium phosphate required = 30 mmol phosphate / (3 mmol / mL)
<u>Volume
potassium phosphate required = 10 mL</u>
This would
also contain potassium in amounts of:
Amount of
potassium in potassium phosphate = 10 mL (4.4 meg / mL)
Amount of
potassium in potassium phosphate = 44 meg
Therefore
the potassium chloride required is:
Volume of
potassium chloride = (80 meg – 44 meg) / (2 meg / mL)
<span><u>Volume of
potassium chloride = 72 mL</u></span>
