Reduction I think... not entirely sure though... when a molecule loses oxygen...?
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
383 mL of 3.50 M muriatic acid solution
Explanation:
First we calculate the number of moles of HCl:
number of moles = mass / molecular weight
number of moles of HCl = 47.7 / 35.5 = 1.34 moles
Now to determine the volume of the 3.50 M muriatic acid (clorhidric acid) solution, we use the following formula:
molar concentration = number of moles / volume (L)
volume (L) = number of moles / molar concentration
volume of the muriatic acid solution = 1.34 / 3.5 = 0.383 L = 383 mL
Learn more about:
molar concentration
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Answer:
moon
Explanation:
the moon's gravitational pull generates tidal force which affects earth's tides
M(H₂O) = 255 g.
m₁(NaCl) = 10.0 g.
m₁(solution) = 255 g + 10 g = 265 g.
ω₁ = 10 g / 265 g · 100%.
ω₁ = 3.77% ÷ 100% = 0.0377.
ω₂= 10% ÷ 100% = 0.1.
ω₂= m₁(NaCl) + m₂(NaCl) / m₁(solution) + m₂(NaCl).
0.1 = 10 g + m₂(NaCl) / 265 g + m₂(NaCl).
26.5 g + 0.1·m₂(NaCl) = 10 g + m₂(NaCl).
0.9·m₂(NaCl) = 16.5 g.
m₂(NaCl) = 18.33 g.
