Matter can not be created nor destroyed. So all of the carbon you begin with you will end up with in the end.
if you start with 1CH4 you have 1mol of C and 4 mol of H you will only be able to make 1mol of CO2 since each molecule of CO2 requires a Carbon atom.
What's most likely going on here is that you are burning CH4
CH4 + 2O2 --> CO2 + 2H2O
the H's end up in water molecules.
According to balanced equation 16g of CH4 gives 44g of CO2.So 1g gives 2.75g of CO2
200 g CH4 x 1 mole CH4/16 g = 12.5 moles CH4
Divide 96 by molecular mass.So the answer is 6mol
156 grams CH4 (1 mole CH4/16.042 grams) = 9.72 moles of methane ==================
c. 16.0 grams CH4/1 mol C04
Methane is non polar.So dispersion forces are formed.
The balanced equation for combustion of CH4 is CH4 + 2O2 ==> CO2 + 2H2OThus, one mole CH4 produces 1 mole CO21 g CH4 x 1 mole CH4/16 g = 0.0625 moles CH40.0625 moles CH4 ==> 0.0625 moles CO20.0625 moles CO2 x 44 g CO2/mole = 2.75 g CO2Thus, the answer would be that 1 grams of CH4 will produce 2.75 grams of CO2 after complete combustion.
1,25 grams of CH4 contain 0,156696.10e23 atoms.
200 g CH4 x 1 mole CH4/16 g = 12.5 moles CH4
Divide 96 by molecular mass.So the answer is 6mol
There are 0.75 moles in it.You have to devide 12 by molecular mass
First, determine molar mass of CH4: C:12g/mol + 4x H:1g/mol= 16g/mol Then divide by the number of grams. 64g/(16g/mol)= 4 moles of CH4
156 grams CH4 (1 mole CH4/16.042 grams) = 9.72 moles of methane ==================
The equation for a complete combustion reaction of CH4 is : CH4 + 2 O2 = CO2 + 2 H2O, showing that one mole of carbon dioxide is formed for each mole of CH4 burned. Therefore, the answer is 44 moles of CO2 formed.
To find the number of moles of Al(OH)3 produced, we need to balance the chemical equation for the reaction where CH4 is formed. Once the equation is balanced, we can determine the mole ratio between Al(OH)3 and CH4. If we assume the balanced equation is: 4CH4 + 3O2 -> 2Al(OH)3 + 4H2O, we can see that the mole ratio is 2:4. Therefore, for every 2 moles of Al(OH)3 produced, 4 moles of CH4 are needed. Since we have 0.560 mol of CH4, we will produce half as many moles of Al(OH)3, which is 0.280 mol. To find the mass in grams, we can multiply the moles by the molar mass of Al(OH)3, which is 78 g/mol. Therefore, 0.280 mol of Al(OH)3 will be produced, which is equal to 21.84 grams.
Two moles of water are produced.
c. 16.0 grams CH4/1 mol C04
CH4 --> CO2 is a 1 to 1 reaction (C-balanced) when burning with oxygen. So 1 mole CH4 --> 1 mole CO2 So 1 Litere CH4 --> 1 Liter CO2 So 16 grams CH4 --> 44 grams CO2