A long chain of hydrocarbon bonded to COOH is a FATTY acid.
Answer:
V = 2.32 Liters
Explanation:
PV = nRT => V = nRT/P
n = 25.8g/122g/mole = 0.21 mole
R = 0.08206 L·atm/mol·K
T = 25.44°C + 273 = 298.44K
P = 2.22 atm (given in problem)
V = (0.21mol)(0.08206 L·atm/mol·K)(298.44K)/(2.22atm) = 2.32 Liters at 25.44°C & 2.22atm
The correct answer is approximately 11.73 grams of sulfuric acid.
The theoretical yield of water from Al(OH)3 is lower than that of H₂SO₄. As a consequence, Al(OH)3 is the limiting reactant, H₂SO₄ is in excess.
The balanced equation is:
2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O
Each mole of Al(OH)3 corresponds to 3/2 moles of H₂SO₄. The molecular mass of Al(OH)3 is 78.003 g/mol. There are 15/78.003 = 0.19230 moles of Al(OH)3 in the five grams of Al(OH)3 available. Al(OH)3 is in limiting, which means that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
The molar mass of H₂SO₄ is 98.706 g/mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.706 = 28.289 g
40 grams of sulfuric acid is available, out of which 28.289 grams is consumed. The remaining 40-28.289 = 11.711 g is in excess, which is closest to the first option, that is, 11.73 grams of H₂SO₄.
Answer:
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Explanation:
<u>1. Chemical balanced equation (given)</u>
<u>2. Mole ratio</u>
This is, 1 mol of NaOH will reacts with 1 mol of KHP.
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<u>3. Find the number of moles in 72.14 mL of the base</u>
<u>4. Find the number of grams of KHP that reacted</u>
The number of moles of KHP that reacted is equal to the number of moles of NaOH, 0.007055 mol
Convert moles to grams:
- mass = number moles × molar mass = 0.007055mol × 204.23g/mol
You have to round to 3 significant figures: 1.44 g (because the molarity is given with 3 significant figures).
<u>5. Find the percentage of KHP in the sample</u>
The percentage is how much of the substance is in 100 parts of the sample.
The formula is:
- % = (mass of substance / mass of sample) × 100
- % = (1.4408g/ 1.864g) × 100 = 77.3%
