<span>Classify is the answer hope this helps :)</span>
Excess reactant : Na
NaCl produced : = 16.497 g
<h3>Further explanation</h3>
Given
Reaction(balanced)
2Na + Cl₂⇒ 2NaCl
20 g Na
10 g Cl₂
Required
Excess reactant
NaCl produced
Solution
mol Na(Ar = 23 g/mol) :
= 20 : 23 = 0.87
mol Cl₂(MW=71 g/mol):
= 10 : 71 g/mol = 0.141
mol : coefficient :
Na = 0.87 : 2 = 0.435
Cl₂ = 0.141 : 1 = 0.141
Limiting reactant : Cl₂(smaller ratio)
Excess reactant : Na
Mol NaCl based on mol Cl₂, so mol NaCl :
= 2/1 x mol Cl₂
= 2/1 x 0.141
= 0.282
Mass NaCl :
= 0.282 x 58.5 g/mol
= 16.497 g
Answer:
4 is the answer
Explanation:
i am not sure for this question
Answer:
2.1 kg of water
Explanation:
Step 1: Given data
- Moles of lithium bromide (solute): 4.3 moles
- Molality of the solution (m): 2.05 m (2.05 mol/kg)
- Mass of water (solvent): ?
Step 2: Calculate the mass of water required
Molality is equal to the moles of solute divided by the kilograms of solvent.
m = moles of solute/kilograms of solvent
kilograms of solvent = moles of solute/m
kilograms of solvent = 4.3 mol /(2.05 mol/kg) = 2.1 kg
