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

Answer 1

Thirty grams of glucose are needed.

Step 1: Begin with the equation that is balanced.

#M_text(r):color(white)(ml) 180.16color(white)(mmmmmmmmmml)44.01# #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"#

#"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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Answer 2

Olivia Anton

The mass of glucose required to produce 44 g of carbon dioxide is 180 g.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.