0.010 moles of carbon c4h10 reacts with oxygon as in the queation 1.76g of carbon dioxide and 0.90 of water are produced. use this information ti work out the balancing numbers for carbon dioxide and water
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Ответ:
The coefficients are 2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O
Step 1. Gather all the information in one place.
M_r: 58.12 32.00 44.01 18.02
aC₄H₁₀ + bO₂ ⟶ cCO₂ + dH₂O
m/g: 1.76 0.90
n/mol: 0.010
Step 2. Calculate the mass of C₄H₁₀.
Mass = 0.010 mol C₄H₁₀ × (58.12 g C₄H₁₀/1 mol C₄H₁₀) = 0.581 g C₄H₁₀
Step 3. Calculate the mass of O₂
Mass of C₄H₁₀ + mass of O₂ = mass of CO₂ + mass of H₂O
0.581 g + x g = 1.76 g + 0.90 g
x = 1.76 + 0.90 - 0.581 = 2.079
Our information now has the form:
M_r: 58.12 32.00 44.01 18.02
aC₄H₁₀ + bO₂ ⟶ cCO₂ + dH₂O
m/g: 0.581 2.079 1.76 0.90
n/mol: 0.010
Step 4. Calculate the moles of each compound.
Moles of O₂ = 2.079 g O₂ × (1 mol O₂/32.00 g O₂) = 0.064 97 mol O₂
Moles of CO₂ = 1.76 g CO₂ × (1 mol CO₂/44.01 g CO₂) = 0.040 00 mol CO₂
Moles of H₂O = 0.90 g H₂O × (1 mol H₂O/18.02 g H₂O) = 0.0499 mol H₂O
Our information now has the form:
aC₄H₁₀ + bO₂ ⟶ cCO₂ + dH₂O
n/mol: 0.010 0.064 97 0.040 00 0.0499
Step 5: Calculate the molar ratios of all the compounds.
a:b:c:d = 0.010:0.064 97:0.040 00:0.0499 = 1:6.497:4.000:4.99
= 2 :12.99:8.00:9.98 ≈ 2:13:8:10
∴ a = 2; b = 13; c = 8; d = 10
The balanced equation is
2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O
Ответ:
4 meters
Step-by-step explanation:
For this question, use trigonometric ratios (sin to be specific). Sin = Opp/Hypotenuse. Thus, we can do sin(30) is equal to 2/hypotenuse. Then, we can simplify to get Hypotenuse = 2/sin(30). Then, we can simplify to get the maximum length as 4 meters.