The balance chemical equation is as follow,
2 Al + 3 O₂ → Al₂O₃
Aluminium is the Limiting Reagent,
As,
107.92 g Al required = 96 g of O₂
Then,
82.49 g of Al will require = X g of O₂
Solving for X,
X = (82.49 g × 96 g) ÷ 107.92 g
X = 73.37 g of O₂
But,
We are provided with 117.65 g of O₂, So, it is provided in excess and 44.28 g of it will remain unreacted.
Solving for Amount of Al₂O₃ formed,
As,
107.92 g of Al produced = 203.92 g of Al₂O₃
Then,
82.49 g of Al will produce = X g of Al₂O₃
SOlving for X,
X = (82.49 g × 203.92 g) ÷ 107.92
X = 155.86 g of Al₂O₃
The maximum number of moles of Nickel Carbonate (NiCO₃) that can form during a precipitation reaction is 0.025 mol. Hope this helped and have a great day!
Energy is used by plants to carry cellular respiration.
Answer:
3.59 moles
Explanation:
Hopefully this helps! :)
Mark as brainliest if right!
Balanced equation:
<span>2 NO + 5 H2 ------> 2 NH3 + 2 H2O
</span>
<span>2 moles NO react with 5 moles H2 to produce 2 moles NH3
</span>
<span>Molar mass of NO = 30.00 g/mol </span>
<span>86.3g NO = 86.3/30.00 = 2.877 moles of NO </span>
<span>This will require: 2.877*5 / 2 = 7.192 moles of H2 </span>
<span>Molar mass of H2 = 2 g/mol </span>
<span>25.6g H2 = 25.6/2 = 12.7 mol H2. </span>
<span>You have excess H2 means the NO is limiting </span>
<span>From the balanced equation: </span>
<span>2 moles of NO will produce 2 moles of NH3 </span>
<span>2.877 moles of NO will produce 2.877 moles of NH3 </span>
<span>Molar mass NH3 = 17g/mol </span>
<span>Mass NH3 produced = 2.877 * 17 = 48.91g
Hence the yield is = 48.91 g ~ 49 g</span>
