Answer:
The answer to your question is 47.44 g of Oxygen
Explanation:
Data
mass of Ammonia = 14.4 g
mass of Oxygen = ?
Balanced chemical reaction
4NH₃ + 7O₂ ⇒ 4NO₂ + 6H₂O
Process
1.- Calculate the molar mass of Ammonia
NH₃ = 4[(1 x 14) + (3 x 1)] = 4[14 + 3] = 4[17] = 68 g
2.- Calculate the molar mass of Oxygen
O₂ = 7[16 x 2] = 7[32] = 224 g
3.- Use proportions to calculate the mass of Oxygen
68g of NH₃ --------------------- 224 g of O₂
14.4 g of NH₃ ----------------- x
x = (14.4 x 224) / 68
x = 3225.6/ 68
x = 47.44 g
Make sure all units are consistent. Convert ml to L first. (500ml = 0.5L). Then divide the mole by the volume which would result to 1.6*10^-4mol/L. This can also be expressed as interms of molarity (M). 1 M = 1 mol/L hence the final answer is <span>1.6*10^-4 M. </span>
Answer:
1) Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make 0.157 mol ammonium sulfate when you neutralize 11.00 g ammonium hydroxide.
2) 2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O.
Explanation:
- Firstly, we should balance the equation of heptane combustion.
- We can balance the equation by applying the conservation of mass to the equation.
- The balanced equation is: <em>2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O.</em>
- This means that every 2.0 moles of ammonium hydroxide (NH₄OH) will produce 1.0 mole of ammonium sulfate (NH₄)₂SO₄ when it is neutralized by sulfuric acid.
- We need to calculate the no. of moles in 11.0 g of ammonium hydroxide that is neutralized using the relation: <em>n = mass/molar mass.
</em>
n of 11.0 g of ammonium hydroxide (NH₄OH) = mass/molar mass = (11.0 g)/(35.04 g/mol) = 0.314 mol.
<u><em>Using cross multiplication:
</em></u>
2.0 moles of NH₄OH make → 1.0 mole of (NH₄)₂SO₄.
0.314 mol of NH₄OH make → ??? moles of (NH₄)₂SO₄.
∴ The no. of moles of (NH₄)₂SO₄ that will be made from neutralizing (11.0 g) of NH₄OH = (0.314 mol)(1.0 mol)/(2.0 mol) = 0.157 mol.
<em>∴ Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make </em><em>0.157</em><em> mol ammonium sulfate when you neutralize 11.00 g ammonium hydroxide.</em>
If enough HCl is not added, the percent yield will be lower as all the magnesium metal has not reacted with HCl.
<h3>What is percent yield of a reaction?</h3>
The percent yield of a reaction is the ratio of the actual yield and the expected yield expressed as a percentage.
- Percent yield = actual yield/expected yield × 100%
The reaction between magnesium metal and HCl liberates hydrogen gas and forms magnesium chloride salt.
If enough HCl was not added to react with all the magnesium metal, the actual yield will be lower.
Hence, percent yield will be lower as well.
Therefore, the percent yield will be lower if all the magnesium metal has not reacted with HCl.
Learn more about percent yield at: brainly.com/question/10046114
