Write formula of compound given two chemicals
1 answer:
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Answer: 0.20 M
Explanation:
According to the dilution law,
where,
= molarity of stock solution = 1.40 M
= volume of stock solution = 72.0 ml
= molarity of diluted solution = m
= volume of diluted solution = 248 ml
Now 124 mL portion of this prepared solution is diluted by adding 133 mL of water.
According to the dilution law,
where,
= molarity of stock solution = 0.41 M
= volume of stock solution = 124 ml
= molarity of diluted solution = m
= volume of diluted solution = (124 +133) ml = 257 ml
Thus the final concentration of the solution is 0.20 M.
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