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The 370 grams of matter were converted into gases and released into the atmosphere during the combustion process. This is due to the fact that wood is composed mainly of carbon, hydrogen, and oxygen, which are transformed into carbon dioxide and water vapor when burned.
For every 1 g of hydrogen burned, 9 g of water is produced. Therefore, if 100 kg (100,000 g) of hydrogen is burned, it will produce 900,000 g (or 900 kg) of water.
To calculate the mass of octane burned, we can use the heat of combustion of octane which is 5470 kJ/mol. First, convert the given energy to kilojoules per mole. Then, use the molar mass of octane to convert moles to grams. This will give you the mass of octane that must be burned.
When 15 liters of propane are completely burned, it produces 110 grams of carbon dioxide.
The mass of H2O produced when propane is burned depends on the amount of propane burned. One mole of propane produces three moles of water. The molar mass of water is 18 g/mol. You can calculate the mass of water produced by multiplying the moles of water by the molar mass.
For every 1 mole of propane burned, 5 moles of oxygen are required. This means that 44 grams of propane requires 160 grams of oxygen to burn completely. Therefore, 100 grams of propane would require (100 grams propane * 160 grams oxygen / 44 grams propane) = 363.64 grams of oxygen to burn completely.
Calculate the mass in grams of water vapor produced if 3.11 moles of propane is burned
The answer is 6,61 g.
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To determine the moles of water produced from the reaction of 6.00 grams of propane, first calculate the moles of propane using its molar mass. Then, use the balanced chemical equation to find the moles of water produced based on the stoichiometry of the reaction.
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To calculate the grams of propane needed to produce 8.00 grams of water, you start by converting the mass of water to moles using the molar mass of water. Then, you use the balanced chemical equation for the combustion of propane to determine the mole ratio between propane and water. Finally, convert the moles of water to moles of propane using the mole ratio and then to grams of propane using the molar mass of propane.
To determine the grams of carbon dioxide produced, we first balance the chemical equation for the combustion of propane: C3H8 + 5O2 -> 3CO2 + 4H2O. Using the stoichiometry of the balanced equation, we find that 42 grams of propane will produce 3 * (44 g/mol) = 132 grams of carbon dioxide. For the water produced, 42 grams of propane will produce 4 * (18 g/mol) = 72 grams of water.
If 100.0g of C3H8 is burned completely, all the carbon in C3H8 will form CO2. The molar mass of C3H8 (propane) is 44.1 g/mol, and it contains 3 carbons. This means 100.0g of C3H8 contains 3*(12.01g) carbon atoms, which will produce 3*(44.01g) of CO2, resulting in 132.03g of CO2 being produced.
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