Answer:
165 g of NaCl are formed in the reaction
Explanation:
2Na + Cl₂ → NaCl
In order to determine the limiting reactant, we convert the mass of each reactant to moles
35 g / 23g/mol = 1.52 moles Na
100 g / 70.9 g/mol = 1.41 moles Cl₂
1 mol of chlorine reacts with 2 moles of Na, so If I have an x value of moles of Cl₂ I would need the double to react.
For 1.41 moles of Cl₂, I need 2.82 moles of Na; therefore my limiting reagent is the Na. Ratio is 2:2. So if I have 2.82 moles of Na I will produce 2.82 moles of NaCl
We convert the moles to mass: 2.82 mol . 58.45 g/1 mol =164.8 g
Answer:
The volume of 2M stock solution added = 2 mL
The volume of water = 198 mL
Explanation:
Considering
Given that:
So,
<u>The volume of 2M stock solution added = 2 mL</u>
<u>The volume of water = 200 - 2 mL = 198 mL</u>
A-Aluminium sulphate.
B-Calcium Chloride.
C-Potassium sulphate.
D-Potassium Nitrate .
E-Calcium Carbonate.
