Atomic Mass of C = 14g/mol
Atomic mass of O = 16g/mol
Molecular mass of CO2 = 12 + 2(16) = 44g/mol
mass = number of moles x molecular mass
mass = 3 mol x 44g/mol = 132g
For this you need the atomic (molecular) mass of CO2. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel. CO2=44.0 grams2.00 moles CO2 × (44.0 grams) = 88.0 grams 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)
1CO2 ==> 1O26.5 g O2 x 1 mol/32 g = 0.203 moles O2grams of CO2 = 0.203 moles x 44 g/mole = 8.9 grams
C + O2 ==> CO2At STP 1 mole CO2 = 22.4 L 10L x 1 mole/22.4 L = 0.446 moles CO2 needed 1 mole C = 1 mole CO2 Therefore you need 0.446 moles C grams C = 0.446 moles C x 12 g/mole = 5.36 grams = 5 g (to 1 significant figure)
1 mole of CO2 has 1 mole of carbon atoms and 2 moles of oxygen atoms. So, 167 mole of CO2 has 167 mole of carbon atoms.
5.0 grams CO2 (1mol CO2/44.01g) = 0.11 moles CO2
6.5 grams CO2 divided by 44 grams CO2 per mole CO2 = 6.8 mole CO2 (Molar mass CO2 = 12 + 2*16 = 44 g/mol)
The answer is 3,99 moles of carbon dioxide.
The answer is 3,99 moles of carbon dioxide.
For this you need the atomic (molecular) mass of CO2. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. CO2=44.0 grams454 grams CO2 / (44.0 grams) = 10.3 moles CO2
For this you need the atomic (molecular) mass of CO2. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel. CO2=44.0 grams2.00 moles CO2 × (44.0 grams) = 88.0 grams CO2
1 mole CO2 has about 44 grams, so half a mole of CO2 equals 22 grams
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)
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.
For this you need the atomic (molecular) mass of CO2. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel. CO2= 44.0 grams1.50 moles CO2 × (44.0 grams) = 66.0 grams CO2
For this you need the atomic (molecular) mass of CO2. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. CO2=44.0 grams33.0 grams C / (44.0 grams) = .750 moles CO2
80,0 moles of CO2is equal to 3 520,8 g.