Answer:
5
Explanation:
all you do is the math expression
The reaction corresponds to the combustion of propane (C3H8). The balanced reaction is:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
The reaction enthalpy is given as:
ΔHrxn = ∑nΔH°f(products) - ∑ nΔH°f(reactants)
= [3ΔH°f(CO2(g)) + 4ΔH°f(H2O(g)] - [1ΔH°f(C3H8(g)) + 5ΔH°f(O2(g)]
= [3(-393.5) + 4(-241.8)] - [-103.9 + 5(0)] = -2043.8 kJ
The enthalpy for the combustion of propane is -2044 kJ
Answer:
Explanation:
<u>1) Mass of carbon (C) in 2.190 g of carbon dioxide (CO₂)</u>
- atomic mass of C: 12.0107 g/mol
- molar mass of CO₂: 44.01 g/mol
- Set a proportion: 12.0107 g of C / 44.01 g of CO₂ = x / 2.190 g of CO₂
x = (12.0107 g of C / 44.01 g of CO₂ ) × 2.190 g of CO₂ = 0.59767 g of C
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<u>2) Mass of hydrogen (H) in 0.930 g of water (H₂O)</u>
- atomic mass of H: 1.00784 g/mol
- molar mass of H₂O: 18.01528 g/mol
- proportion: 2 × 1.00784 g of H / 18.01528 g of H₂O = x / 0.930 g of H₂O
x = ( 2 × 1.00784 g of H / 18.01528 g of H₂O) × 0.930 g of H₂O = 0.10406 g of H
<u>3) Mass of oxygen (O) in 1.0857 g of pure sample</u>
- Mass of O = mass of pure sample - mass of C - mass of H
- Mass of O = 1.0857 g - 0.59767 g - 0.10406 = 0.38397 g O
Round to four decimals: Mass of O = 0.3840 g
<u>4) Mole calculations</u>
Divide the mass in grams of each element by its atomic mass:
- C: 0.59767 g / 12.0107 g/mol = 0.04976 mol
- H: 0.10406 g / 1.00784 g/mol = 0.10325 mol
- O: 0.3840 g / 15.999 g/mol = 0.02400 mol
<u>5) Divide every amount by the smallest value (to find the mole ratios)</u>
- C: 0.04976 mol / 0.02400 mol = 2.07 ≈ 2
- H: 0.10325 mol / 0.02400 mol = 4.3 ≈ 4
- O: 0.02400 mol / 0.02400 mol = 1
Thus the mole ratio is 2 : 4 : 1, and the empirical formula is:
Answer:
magnesium ion Hg2+ cation is formed
