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Consider the combustion reaction of acetylene (C_(2)H_(2)) 2C_(2)H_(2)+5O_(2)arrow 4CO_(2)+2H_(2)O Use the periodic table to determine how many grams of oxygen would be required to react completely with 859.0gC_(2)H_(2) 423.0gO_(2) 832.0gO_(2) 1,750.gO_(2) 2,640.gO_(2)

Question

Consider the combustion reaction of acetylene (C_(2)H_(2))
2C_(2)H_(2)+5O_(2)arrow 4CO_(2)+2H_(2)O
Use the periodic table to determine how many grams of oxygen would be required to react completely
with 859.0gC_(2)H_(2)
423.0gO_(2)
832.0gO_(2)
1,750.gO_(2)
2,640.gO_(2)

Consider the combustion reaction of acetylene (C_(2)H_(2)) 2C_(2)H_(2)+5O_(2)arrow 4CO_(2)+2H_(2)O Use the periodic table to determine how many grams of oxygen would be required to react completely with 859.0gC_(2)H_(2) 423.0gO_(2) 832.0gO_(2) 1,750.gO_(2) 2,640.gO_(2)

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FreddieElite · Tutor for 8 years

Answer

\(2,640 \text{ g of } \mathrm{O}_{2}\).

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To solve this problem, we need to follow these steps:1. Determine the molar mass of acetylene (\(\mathrm{C}_{2} \mathrm{H}_{2}\)).2. Convert the mass of acetylene to moles.3. Use the balanced chemical equation to find the mole ratio between acetylene and oxygen.4. Convert the moles of acetylene to moles of oxygen using the mole ratio.5. Determine the molar mass of oxygen (\(\mathrm{O}_{2}\)).6. Convert the moles of oxygen to grams.Let's go through each step:Step 1: Determine the molar mass of acetylene (\(\mathrm{C}_{2} \mathrm{H}_{2}\)).The molar mass of carbon (C) is approximately 12.01 g/mol, and the molar mass of hydrogen (H) is approximately 1.008 g/mol. Acetylene has 2 carbon atoms and 2 hydrogen atoms.Molar mass of \(\mathrm{C}_{2} \mathrm{H}_{2}\) = \(2 \times 12.01\) g/mol + \(2 \times 1.008\) g/mol = 24.02 g/mol + 2.016 g/mol = 26.036 g/mol.Step 2: Convert the mass of acetylene to moles.Using the molar mass of acetylene, we can convert the given mass of acetylene to moles:\(\text{moles of } \mathrm{C}_{2} \mathrm{H}_{2} = \frac{859.0 \text{ g}}{26.036 \text{ g/mol}}\).Step 3: Use the balanced chemical equation to find the mole ratio between acetylene and oxygen.From the balanced equation, 2 moles of \(\mathrm{C}_{2} \mathrm{H}_{2}\) react with 5 moles of \(\mathrm{O}_{2}\).Step 4: Convert the moles of acetylene to moles of oxygen using the mole ratio.\(\text{moles of } \mathrm{O}_{2} = \text{moles of } \mathrm{C}_{2} \mathrm{H}_{2} \times \frac{5 \text{ moles of } \mathrm{O}_{2}}{2 \text{ moles of } \mathrm{C}_{2} \mathrm{H}_{2}}\).Step 5: Determine the molar mass of oxygen (\(\mathrm{O}_{2}\)).The molar mass of oxygen is approximately 16.00 g/mol. Since \(\mathrm{O}_{2}\) has two oxygen atoms, the molar mass is \(2 \times 16.00\) g/mol = 32.00 g/mol.Step 6: Convert the moles of oxygen to grams.\(\text{grams of } \mathrm{O}_{2} = \text{moles of } \mathrm{O}_{2} \times 32.00 \text{ g/mol}\).Now, let's do the calculations:\(\text{moles of } \mathrm{C}_{2} \mathrm{H}_{2} = \frac{859.0 \text{ g}}{26.036 \text{ g/mol}} = 32.99 \text{ moles}\) (rounded to two decimal places for intermediate calculation).\(\text{moles of } \mathrm{O}_{2} = 32.99 \text{ moles} \times \frac{5}{2} = 82.475 \text{ moles}\).\(\text{grams of } \mathrm{O}_{2} = 82.475 \text{ moles} \times 32.00 \text{ g/mol} = 2639.2 \text{ g}\).Rounded to three significant digits, the answer is \(2640 \text{ g of } \mathrm{O}_{2}\).
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