Answer:
See below.
Explanation:
The mass of octane in the sample of gasoline is 0.02851 * 482.6 = 13.759 g of octane.
The balanced equation is:
2C8H18(l) + 25O2(g) ----> 16CO2(g) + 18H2O(g)
From the equation, using atomic masses:
228.29 g of octane forms 704 g of CO2 and 324.3 g of H2O
So the mass of CO2 formed from the combustion of 13.759 g of octane = (704 * 13.759) / 228.29
= 42.43 g of CO2.
Amount of water = 324.3 * 13.759) / 228.29
= 19.55 g of H2O.
The correct answer is that 1.125 mol of NaOH is available, and 60.75 g of FeCl₃ can be consumed.
The mass of NaOH is 45 g
The molar mass of NaOH = 40 g/mol
The moles of NaOH = mass / molar mass
= 45 / 40
= 1.125
Thus, 1.125 mol NaOH is available
3 NaOH + FeCl₃ ⇒ Fe (OH)₃ + 3NaCl
3 mol of NaOH react with 1 mol of FeCl₃
1.125 moles of NaOH will react with x moles of FeCl₃
x = 1.125 / 3
x = 0.375 mol
0.375 mol FeCl₃ can take part in reaction
The molar mass of FeCl₃ is 162 g/mol
The mass of FeCl₃ = moles × mass
= 0.375 × 162
= 60.75 g
Thus, the amount of FeCl₃, which can be consumed is 60.75 g
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