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The number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g
First, we will write the balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
Ba(OH)₂ + 2NH₄NO₃ → 2NH₄OH + Ba(NO₃)₂
This means, 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate
Now, we will calculate the number of moles of barium hydroxide present.
Mass of barium hydroxide (Ba(OH)₂) = 20 g
Using the formula
Molar mass of Ba(OH)₂ = 171.34 g/mol
∴ Number of moles of Ba(OH)₂ present =
Number of moles of Ba(OH)₂ present = 0.116727 mole
Now,
Since 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate
Then,
0.116727 mole of barium hydroxide will react with 2 × 0.116727 mole of ammonium nitrate
2 × 0.116727 = 0.233454 mole
∴ Number of moles of NH₄NO₃ required is 0.233454 mole
Now, for the mass of ammonium nitrate (NH₄NO₃) required
From the formula
Mass = Number of moles × Molar mass
Molar mass of NH₄NO₃ = 80.043 g/mol
∴ Mass of NH₄NO₃ required = 0.233454 × 80.043
Mass of NH₄NO₃ required = 18.68636 g
Mass of NH₄NO₃ required ≅ 18.7g
Hence, the number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g
Learn more on determining mass of reactant required here: brainly.com/question/11232389
First a balanced reaction equation must be established:
→ 
Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
= 4.83 mols
Now the mole ratio of Al : O₂ based on the equation is 4 : 3
[4Al + 3 O₂ → 2 Al₂O₃]
∴ if moles of Al = 4.83 moles
then moles of O₂ = (4.83 mol ÷ 4) × 3
= 3.63 mol (to 2 sig. fig.)
Thus it can be concluded that 3.63 moles of oxygen is needed to react completely with 145 g of aluminum.
Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
= 4.83 mols
Now the mole ratio of Al : O₂ based on the equation is 4 : 3
[4Al + 3 O₂ → 2 Al₂O₃]
∴ if moles of Al = 4.83 moles
then moles of O₂ = (4.83 mol ÷ 4) × 3
= 3.63 mol (to 2 sig. fig.)
Thus it can be concluded that 3.63 moles of oxygen is needed to react completely with 145 g of aluminum.