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2 C8H18(l) + 25 O2(g)----->16 CO2(g) + 18 H2O(l)
(a) How many grams of O2 are needed to burn 1.90 g of C8H18?
=_______g
(b) Octane has a density of 0.692 g/mL at 20°C. How many grams of O2 are required to burn 8.40 gal. of C8H18?
=________g
Answer:
So we have 0.0167 moles of octane and we consume 2 moles of octane for every 25 moles of oxygen gas produces
0.0167 moles of C8H18 * (25 moles O2/2 moles C8H18) = 0.208 moles of O2
but we want grams, so 0.208 moles O2 * (32.00 g/mol of O2) = 6.67 grams of O2
Part 2 -
We've got 8.40 gal of octane, there are 3.785 L/gal so
8.40 gal of C8H18 * 3.786 (L/gal) = 32.25 L of C8H18
32.25 L = 32,550 mL
32,550 mL * density of octane (0.692 g/mL) = 22,524 g of octane = 197.6 moles of octane
therefore, we consume this number of moles of O2: 197.6 mol C8H18 * (25 mol O2/2 mol C8H18) =2469.8 mol O2 * (32 g of O2/mol of O2) = 79,033 g of O2.
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