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[ 34.3(gC3H8) / 44(g/mol C3H8)] * [ 3mol CO2 / 1molC3H8 ] * 44(g/mol CO2) = 103 gram CO2
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
The empirical formula for aluminum chloride is AlCl3, and its gram formula mass is 133.34. The formula shows that each formula unit contains one aluminum atom, and the the gram atomic mass of aluminum is 26.9815. Therefore, 18(133.34/26.9815) or 89 grams, to the justified number of significant digits, of aluminum chloride will be produced.
5 grams
a gram is a gram no mater what
[ 34.3(gC3H8) / 44(g/mol C3H8)] * [ 3mol CO2 / 1molC3H8 ] * 44(g/mol CO2) = 103 gram CO2
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
There is 1 gram per gram.
no
Gram
The plural of gram is grams.
1 hector gram = 100 grams
grams but
0.001 gram. There are 1000mg in a gram
0.001 gram. There are 1000mg in a gram
1 kg = 1000 grams 1 gram = 0.001 kg
The equation for the reaction is C + O2 -> CO2. The relevant gram atomic masses are 12.011 for carbon and 15.9994 for oxygen. Therefore, the ratio of the mass of carbon dioxide produced to carbon burnt is [2(15.9994) + 12.011]/12.011 or about 3.66. From burning 3 grams of carbon, the mass of carbon dioxide produced is therefore 1 X 101 grams, to the justified number of significant digits.