Multiply the value of 1 mole by 5.
44 g/mol x 5 mol
Mole units cancel out, leaving 220g of CO2
First write the balance equation: Na2CO3 + 2HNO3 ==> 2NaNO3 + CO2 + H2O Next calculate moles of Na2CO3 used: 7.5 g x 1 mole/106 g = 0.071 moles Na2CO3 Then look at mole ratio of Na2CO3 to CO2 and see that it is 1 to 1 Thus, moles CO2 produced = 0.071 moles Finally, convert moles CO2 to grams of CO2: 0.071 moles x 44g/mole = 3.1 g (to 2 significant figures)
Carbon dioxide is CO2 and has a molar mass of 44g whereas propane is C3H8 and has a molar mass of 44g as well. Therefore, both are of equal mass
to work out the number of moles, you divide the mass by the relative atomic mass so 200 divided by 44 = 4.55 moles
Molecular weight of CO2 is about 44g (12g Carbon + 32g Oxygen *2 atoms Oxygen). So that means if you are presented with 44g CO2, that means there's 12g Carbon.
Atomic mass of C = 14g/mol Atomis mass of O = 16g/mol Molecular mass of CO2 = 12 + 2(16) = 44g/mol number of moles = mass / molecular mass number of moles = 220g / 44g/mol = 5.0mol 1 mole of substance contains approximately 6.02 x 10^23 of its particles 5.0 moles of CO2 have 5.0 x 6.02 x 10^23 = 3.01 x 10^24 CO2 molecules. Each CO2 molecule contains 2 oxygen atoms. Total number of Oxygen atom = 2 x 3.01 x 10^24 = 6.02 x 10^24 atoms
The molar mass of carbon dioxide (CO2) is 44 g/mol. Carbon has a molar mass of 12 g/mol and oxygen has a molar mass of 16 g/mol. So, in 44g of CO2, there are 12g of carbon. Therefore, the percent by mass of carbon in 44g of carbon dioxide is (12g / 44g) x 100 = 27.3%.
First you need to work out the balanced equation. 2C2H2 + 5O2 -----> 4CO2 + 2H2O From this we see that 2 moles of acetylene produces 4 moles of carbon dioxide. 1 mole of carbon dioxide is 12 + 16 + 16 = 44g (adding the mass numbers of the component elements). If 1 mole = 44g then 4 mole = 176g
44g/mol
So CO2 is 44g/mol. 48/44 is 1.0909091 moles. This multiplied by avagadros number is 6.57x1023 molecules. As there is 1 carbon per molecule, this also equals the number of carbon atoms
First write the balance equation: Na2CO3 + 2HNO3 ==> 2NaNO3 + CO2 + H2O Next calculate moles of Na2CO3 used: 7.5 g x 1 mole/106 g = 0.071 moles Na2CO3 Then look at mole ratio of Na2CO3 to CO2 and see that it is 1 to 1 Thus, moles CO2 produced = 0.071 moles Finally, convert moles CO2 to grams of CO2: 0.071 moles x 44g/mole = 3.1 g (to 2 significant figures)
molecular weight of carbon dioxide = 44u gram molecular weight of carbon dioxide = 44g given mass = 11g no. of moles = given mass divided by molar mass = 11 divided by 44 = 0.25moles.
To determine the grams of calcium carbonate needed, we first calculate the moles of carbon dioxide using the ideal gas law. At STP, 1 mole of any ideal gas occupies 22.4 L. Therefore, 49.0 L of carbon dioxide is 49.0/22.4 moles. From the balanced chemical equation, we know that 1 mole of calcium carbonate produces 1 mole of carbon dioxide. Finally, using the molar mass of calcium carbonate, we can convert moles to grams.
Carbon dioxide is CO2 and has a molar mass of 44g whereas propane is C3H8 and has a molar mass of 44g as well. Therefore, both are of equal mass
to work out the number of moles, you divide the mass by the relative atomic mass so 200 divided by 44 = 4.55 moles
Molecular weight of CO2 is about 44g (12g Carbon + 32g Oxygen *2 atoms Oxygen). So that means if you are presented with 44g CO2, that means there's 12g Carbon.
C + O2 -------> CO2 12g of carbon produces 44g of carbon dioxide 1kg of carbon will produce 3-67kg of carbon dioxide
In the combustion of C7H16, 7 moles of O2 are required to form 8 moles of CO2. From the given 19.4g of CO2, 1 mole of CO2 is equivalent to 44g. Using stoichiometry, we can calculate that 25.23g of O2 are consumed during the combustion.