4. 50.02 g of (NH₄)₂S
5. 7.12 g of Ga₂O₃
6. 55.07 g of NiCl₂
7. 30.6 %
Thanksful, you have the answers!
Explanation:
4. Reaction: 2NH₃ + H₂S → (NH₄)₂S
We determine the limiting reagent. For that propose we convert the mass of reactants to moles:
25 g / 17 g/mol = 1.47 moles of ammonia
96 g / 34.06 g/mol = 2.82 moles of sulfur
1 mol of sulfur hydrogen reacts with 2 moles of ammonia
2.82 moles of H₂S will react with (2.82 . 2) /1 = 5.64 moles of ammonia.
Clearly, the NH₃ is the limiting reactant.
2 moles of ammonia produce 1 mol of ammonium sulfur
Then, 1.74 moles of ammonia will produce (1.47 . 1) / 2 = 0.735 moles of sulfur. We convert the moles to mass:
0.735 mol . 68.06 g / 1mol = 50.02 g
5. Word equation:
Gallium reacts with oxygen in order to produce gallium oxide.
Formula equation: 4Ga + 3O₂ → 2Ga₂O₃
We convert the mass of reactants to moles:
5.3 g / 69.72 g/mol = 0.0760 moles of Ga
5.3 g / 32 g/mol = 0.166 moles of O₂
3 moles of oxygen react with 4 moles of Ga
Therefore, 0.166 moles of O₂ will react with (0.166 . 4) / 3 = 0.221 moles of Ga. As we do not have the amount required, Ga is the limiting reactant.
4 moles of Ga, can produce 2 moles of gallium oxide
0.0760 moles of Ga will produce (0.0760 . 2) /4 = 0.0380 moles of Ga₂O₃
We convert the moles to mass: 0.0380 mol . 187.44 g / 1 mol = 7.12 g of Ga₂O₃
6. Reaction is: 3CaCl₂ + Ni₃(AsO₄)₂ → 3 NiCl₂ + Ca₃(AsO₄)₂
47.2 g / 110.98 g/mol = 0.425 moles of chloride
135.2 g / 453.91 g /mol = 0.298 moles of Ni₃(AsO₄)₂
1 mol of Ni₃(AsO₄)₂ reacts with 3 moles of calcium chloride
Then, 0.298 moles will react with (0.298 . 3)/ 1 = 1.27 moles of CaCl₂
Clearly, the CaCl₂ is the limiting reactant.
Ratio is 3:3, 3 moles of CaCl₂ can produce 3 moles of NiCl₂. Therefore 0.425 moles will produce the same amount of NiCl₂.
We convert the moles to mass: 0.425 mol. 129.59 g /1mol = 55.07 g of NiCl₂
7. Let's determine the reaction:
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
As the oxygen is in excess, the limiting reagent is the ethene. We convert the mass to moles → 170.9 g / 28 g/mol = 6.10 moles
So, in the theoretical yield 1 mol of ethene can produce 2 moles of CO₂
Therefore 6.10 moles will produce (6.10 . 2) /1 = 12.20 moles of CO₂
The mass produced is 536.8 g (12.20 mol . 44 g /1mol)
In order to determine the percent yield we have to apply this formula
(Produced yield/Theoretical yield) . 100
We replace data: (164.1 g / 536.8 g) .100 = 30.6 %
