19.03.2023 - 09:12

Pentane gas (C5H12) reacts with oxygen to produce carbon dioxide and water. C5H12(g) + 8O2(g) arrow 5CO2(g) + 6H2O(g) How many grams of H2O are formed if 44.6 grams of C5H12 is mixed with 104 grams of O2?

Question:

Pentane gas {eq}(C_5H_{12}) {/eq} reacts with oxygen to produce carbon dioxide and water.

{eq}C_5H_{12}(g) + 8O_2(g) to 5CO_2(g) + 6H_2O(g) {/eq}

How many grams of {eq}H_2O {/eq} are formed if 44.6 grams of {eq}C_5H_{12} {/eq} is mixed with 104 grams of {eq}O_2 {/eq}?

Answers (1)
  • Elouise
    April 12, 2023 в 01:10
    We will use stoichiometry to solve this problem. The balanced chemical equation shows that 1 mol of pentane reacts with 8 mol of oxygen to produce 6 mol of water. Step 1: Calculate the number of moles of pentane present: 44.6 g of C5H12 = (44.6 g) / (72.15 g/mol) = 0.618 mol Step 2: Calculate the number of moles of oxygen present: 104 g of O2 = (104 g) / (32 g/mol) = 3.25 mol Step 3: Determine the limiting reactant: To determine the limiting reactant, we compare the number of moles of pentane and oxygen. According to the balanced chemical equation, 1 mol of pentane requires 8 mol of oxygen. Therefore, to react completely, 0.618 mol of pentane requires (8 mol O2/mol C5H12) x 0.618 mol C5H12 = 4.944 mol of oxygen. Since only 3.25 mol of oxygen are available, oxygen is the limiting reactant. Step 4: Calculate the number of moles of water produced: From the balanced chemical equation, 1 mol of pentane produces 6 mol of water. So, 0.618 mol of pentane produces (6 mol H2O/mol C5H12) x 0.618 mol C5H12 = 3.708 mol of water. Step 5: Calculate the mass of water produced: The molecular weight of water is 18 g/mol. Therefore, the mass of water produced is: (3.708 mol) x (18 g/mol) = 66.744 g Therefore, 66.744 grams of H2O are formed if 44.6 grams of C5H12 is mixed with 104 grams of O2.
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