In order to find the molarity of the solution, we first require the moles of acetic acid added. For this,we need the mass which is:
Mass = volume * density
Mass = 50 * 1.05
Mass = 52.5 grams
Moles = mass / molecular weight
Moles = 52.5 / 60.05
Moles = 0.874 mol
Next, we know that the molarity of a solution is:
Molarity = moles / liter
Molarity = 0.874 / 0.5
Molarity = 1.75 M
1 is hours and 2 is meters
Answer:
4.37 g of barium sulphate
Explanation:
The reaction equation is;
3BaCl2(aq) + Fe2(SO4)3(aq) ---->3 BaSO4(s) + 2FeCl3(aq)
From the question, the number of moles of both barium chloride and FeSO4 = 125/1000 L × 0.150 M = 0.01875 moles
To find the limiting reactant;
3 moles of barium chloride yields 3 moles of barium sulphate
0.01875 moles of barium chloride yields 3 × 0.01875 moles/3 = 0.01875 moles of barium sulphate
1 mole of iron III sulphate yields 3 moles of barium sulphate
0.01875 molesof iron III sulphate yields 0.01875 moles ×3/1 = 0.05625 moles of barium sulphate
Hence,barium chloride is the limiting reactant
Amount of barium sulphate produced = 0.01875 moles × 233 g/mol = 4.37 g of barium sulphate
Answer:
3.68 grams.
Explanation:
First we <u>convert 9.5 g of NaCl into moles of NaCl</u>, using its<em> molar mass</em>:
9.5 g ÷ 58.44 g/mol = 0.16 mol NaCl
In<em> 0.16 moles of NaCl there are 0.16 moles of sodium </em>as well.
We now <u>convert 0.16 moles of sodium into grams</u>, using <em>sodium's molar mass</em>:
0.16 mol * 23 g/mol = 3.68 g
