The combustion of glucose $C_{6} H_{12} O_{6}$ $"C"_6"H"_12"O"_6$ produces carbon dioxide and water. What mass of glucose will produce 44 g of carbon dioxide?

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The combustion of glucose $C_{6} H_{12} O_{6}$ $"C"_6"H"_12"O"_6$ produces carbon dioxide and water. What mass of glucose will produce 44 g of carbon dioxide?

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Answer 1 · 2017-09-05T00:02:56

The mass of glucose required is 30 g.

Step 1. Start with the balanced equation

$M_{r} : m l 180.16 m m m m m m m m m m l 44.01$ $M_text(r):color(white)(ml) 180.16color(white)(mmmmmmmmmml)44.01$
$m m m C_{6} H_{12} O_{6} + {6O}_{2} \to {6CO}_{2} + 6 H_{2} O$ $color(white)(mmm)"C"_6"H"_12"O"_6 + "6O"_2 → "6CO"_2 + 6"H"_2"O""$

Step 2. Calculate the moles of ${CO}_{2}$ $"CO"_2"$

$"Moles of CO"_2 = 44 color(red)(cancel(color(black)("g CO"_2))) × ("1 mol CO"_2)/(44.01color(red)(cancel(color(black)("g CO"_2)))) = "1.00 mol CO"_2"$

Step 3. Calculate the moles of $"C"_6"H"_12"O"_6$

$"Moles of C"_6"H"_12"O"_6 = 1.00 color(red)(cancel(color(black)("mol CO"_2))) × (1 "mol C"_6"H"_12"O"_6)/(6color(red)(cancel(color(black)("mol CO"_2)))) = "0.167 mol C"_6"H"_12"O"_6$

Step 4. Calculate the mass of $"C"_6"H"_12"O"_6$

$"Mass of C"_6"H"_12"O"_6 = 0.167 color(red)(cancel(color(black)("mol C"_6"H"_12"O"_6))) × ("180.16 g C"_6"H"_12"O"_6)/(1 color(red)(cancel(color(black)("mol C"_6"H"_12"O"_6)))) = "30 g C"_6"H"_12"O"_6$

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The combustion of glucose $C_{6} H_{12} O_{6}$ $"C"_6"H"_12"O"_6$ produces carbon dioxide and water. What mass of glucose will produce 44 g of carbon dioxide?

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The combustion of glucose "C"_6"H"_12"O"_6C6H12O6"C"_6"H"_12"O"_6 produces carbon dioxide and water. What mass of glucose will produce 44 g of carbon dioxide?

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The combustion of glucose $C_{6} H_{12} O_{6}$ $"C"_6"H"_12"O"_6$ produces carbon dioxide and water. What mass of glucose will produce 44 g of carbon dioxide?