Sulphur trioxide reacts with water to form a solution of sulphuric acid.
The equation describing this reaction is:
SO3 + H2O .............> H2SO4
Based on the above equation, the formula of the compound formed when sulphur trioxide reacts with water is: H2SO4
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:
the answer is O²- hopefully
The equilibrium constant expression is the ratio of the concentrations of the products over the reactants. Notice how each concentration of product or reactant is raised to the power of its coefficient. For example, the concentration of D is raised to the power of 3 since it is 3D in the balanced reaction.
The number of protons is equal to their atomic number ever time.
For example: Carbon always has 6 protons
And the number of neutrons depends on whether it is Carbon 12 or 13 or 14 etc.
