A 56.8g sample of aluminum is heated from 79.5°C to 143.7°C. The specific heat capacity of aluminum is 0.900 J/(g*K). Calculate
1 answer:
You might be interested in
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