Answers:
8.70 g
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s <em>gather all the information</em> in one place.
M_r: 32.00 44.01
2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 18H₂O
m/g: 9.88
(a) Calculate the <em>moles of O₂
</em>
n = 9.88 g O₂ ×1 mol O₂ /32.00 g O₂
n = 0.3088 mol O₂
(b) Calculate the <em>moles of CO₂</em>
The molar ratio is (16 mol CO₂/25 mol O₂)
n = 0.3088 mol O₂ × (16 mol CO₂/25 mol O₂)
n = 0.1976 mol CO₂
(c) Calculate the <em>mass of CO₂
</em>
Mass of CO₂ = 0.1976 mol CO₂ × (44.01 g CO₂/1 mol CO₂)
Mass of CO₂ = 8.70 g CO₂
Answer:
2n^2 electrons
Explanation:
for example, the first level contain 2*(1^2) = 2 electrons, the second level contain 2*(2^2) = 8 electrons, etc
True because if anything is moving it is in motion. And because horizontal is similar to projectile!
Answer:
Mass = 2.89 g
Explanation:
Given data:
Mass of NH₄Cl = 8.939 g
Mass of Ca(OH)₂ = 7.48 g
Mass of ammonia produced = ?
Solution:
2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2NH₃ + 2H₂O
Number of moles of NH₄Cl:
Number of moles = mass/molar mass
Number of moles = 8.939 g / 53.5 g/mol
Number of moles = 0.17 mol
Number of moles of Ca(OH)₂ :
Number of moles = mass/molar mass
Number of moles = 7.48 g / 74.1 g/mol
Number of moles = 0.10 mol
Now we will compare the moles of ammonia with both reactant.
NH₄Cl : NH₃
2 : 2
0.17 : 0.17
Ca(OH)₂ : NH₃
1 : 2
0.10 : 2/1×0.10 = 0.2 mol
Less number of moles of ammonia are produced by ammonium chloride it will act as limiting reactant.
Mass of ammonia:
Mass = number of moles × molar mass
Mass = 0.17 mol × 17 g/mol
Mass = 2.89 g
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