Answer:
53.6 g of N₂H₄
Explanation:
The begining is in the reaction:
N₂(g) + 2H₂(g) → N₂H₄(l)
We determine the moles of each reactant:
59.20 g / 28.01 g/mol = 2.11 moles of nitrogen
6.750 g / 2.016 g/mol = 3.35 moles of H₂
1 mol of N₂ react to 2 moles of H₂
Our 2.11 moles of N₂ may react to (2.11 . 2) /1 = 4.22 moles of H₂, but we only have 3.35 moles. The hydrogen is the limiting reactant.
2 moles of H₂ produce at 100 % yield, 1 mol of hydrazine
Then, 3.35 moles, may produce (3.35 . 1)/2 = 1.67 moles of N₂H₄
Let's convert the moles to mass:
1.67 mol . 32.05 g/mol = 53.6 g
Answer:
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Explanation:
You can use M x V = M' x V'
3 x V = 250 x 1.2
V = 100 ml
Balanced:
1. <span>Na2O + H2O ---> 2NaOH
2. </span><span>K2O + H2O ---> 2KOH
3. </span><span>MgO + H2O ---> Mg(OH)2
4. </span><span>CaO + H2O ---> Ca(OH)2
5. </span><span>SO2 + H2O ⇄ H2SO3
6. </span>SO3 + H2O ---> H2SO4
All except by 2 were balanced.
