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5.0 grams CO2 (1mol CO2/44.01g) = 0.11 moles CO2
C + O2 ------> CO2 is the reaction equation. 1 mole of C + 1 mole of O2 makes 1 mole of CO2.
6.5 grams CO2 divided by 44 grams CO2 per mole CO2 = 6.8 mole CO2 (Molar mass CO2 = 12 + 2*16 = 44 g/mol)
The number is zero.No CO2 is produced in glycolisis.
92.4 grams
CaC12
5.0 grams CO2 (1mol CO2/44.01g) = 0.11 moles CO2
If all of th 120 g of glucose are converted to energy, how many grams of h2o and co2 will be produced?
1.3 mole C2H2 will produce 2.6 mole CO2, weighting 2.6(mol) x 44 (g/mol) = 114 g CO2
228.8 g of CO2 228.8 g
C6H12O6 + 6O2 --> 6CO2 + 6H2O 45 grams C6H12O6 (1 mole C6H12O6/180.156 grams)(6 moles CO2/1 mole C6H12O6)(44.01 grams/1 mole CO2) = 66 grams carbon dioxide produced ==========================
[ 34.3(gC3H8) / 44(g/mol C3H8)] * [ 3mol CO2 / 1molC3H8 ] * 44(g/mol CO2) = 103 gram CO2
The grams of Carbon present in C12H22O11 depends on how many grams of C12H22O11 you have. For every 342 grams of C12H22O11 that you have, you will have 12 g of carbon.
C2H4 + 3O2 -> 2CO2 +2H2O 50.0 grams C2H4 (1 mole C2H4/28.052 grams)(2 mole CO2/1 mole C2H4)(44.01 grams/1 mole CO2) = 157 grams CO2 produced
C + O2 ------> CO2 is the reaction equation. 1 mole of C + 1 mole of O2 makes 1 mole of CO2.
6.5 grams CO2 divided by 44 grams CO2 per mole CO2 = 6.8 mole CO2 (Molar mass CO2 = 12 + 2*16 = 44 g/mol)
1CO2 ==> 1O26.5 g O2 x 1 mol/32 g = 0.203 moles O2grams of CO2 = 0.203 moles x 44 g/mole = 8.9 grams