Why Sncl2 is solid but SnCl4 is liquid at room temperature
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Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ
Explanation :
First we have to calculate the moles of n-butane.
Given:
Molar mass of n-butane = 58.12 g/mole
Mass of n-butane = 58.3 g
Now put all the given values in the above expression, we get:
Now we have to calculate the energy required.
where,
Q = energy required
= enthalpy of fusion of solid n-butane = 4.66 kJ/mol
n = moles = 1.00 mol
Now put all the given values in the above expression, we get:
Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ
N₂ : limiting reactant
H₂ : excess reactant
<h3>Further explanation</h3>Given
mass of N₂ = 100 g
mass of H₂ = 100 g
Required
Limiting reactant
Excess reactant
Solution
Reaction
<em>N₂+3H₂⇒2NH₃</em>
mol N₂(MW=28 g/mol) :
mol H₂(MW= 2 g/mol) :
A method that can be used to find limiting reactants is to divide the number of moles of known substances by their respective coefficients, and small or exhausted reactans become a limiting reactants
From the equation, mol ratio N₂ : H₂ = 1 : 3, so :
N₂ becomes a limiting reactant (smaller ratio) and H₂ is the excess reactant