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9 particles
it was burned off because of energy forming heat or fire.
1.828e25
44
Glucose
Calculate the mass in grams of water vapor produced if 3.11 moles of propane is burned
Let's see. C3H8 + 5O2 -> 3CO2 + 4H2O For every one molecule of propane burned there is four molecules of water produced. Or, this is the actuality. 1 molecule propane (1 mole C3H8/6.022 X 1023)(4 mole H2O/ 1mole C3O8) = 6.64 X 10 -24 molecules water
If 15 liters of propane are completely consumed 90,25 grams of carbon dioxide are produced.
The answer is 6,61 g.
7
9 particles
If all of th 120 g of glucose are converted to energy, how many grams of h2o and co2 will be produced?
Assuming that the water is produced by complete combustion of propane, the balanced equation is: C3H8 + 5 O2 = 3 CO2 + 4 H2O. The gram molecular mass of propane is three times the gram atomic mass of carbon plus eight times the gram atomic mass of hydrogen = 3(12.011) + 8(1.008) = 44.097, and the gram molecular mass of water is the gram atomic mass of oxygen plus twice the gram atomic mass of hydrogen = 15.999 + 2(1.008) = 18.015. The balanced reaction equation shows that four molecules of water are produced for each molecule of propane, so that the ratio of grams of propane reacted to grams of water produced is 44.097/(4 X 18.015) = 0.61191, to the justified number of significant digits. Therefore, the amount of propane required to produce 8.00 grams of water = 8.00(0.61191) = 4.90 grams to the clearly justified number of significant digits, matching the three significant digits given for 8.00, or 4.895 grams, where the last digit is depressed because it may well not be accurate within + 1
it was burned off because of energy forming heat or fire.
The reaction isC3H8 + 5O2 ----> 3CO2 + 4H2O100g of propane is approx 2.27 moles.From the equation above, we see that the ratio of C3H8 to CO2 is 1:3, therefore the number of moles of CO2 which form is approx 6.82.This relates to a mass of 300g of CO2
322 grams.
96.75 grams of NaCl