2C2H6 +7O2 --> 4CO2 + 6H2O
If 6 moles of O2 are used in the following reaction, how many grams of CO2 will be produced?
2C2H6 +7O2 --> 4CO2 + 6H2O
since this is already a balanced equation, we can now use their stoich coefficients for the ratios.
first, since we need grams of CO2, we find it's molecular mass. from the periodic table, the masses of C and O are
C = 12 and O = 16 , thus
1*12 + 2*16 = 44 g/mol
therefore,
6 mol O2 * (4 mol CO2 / 7 mol O2) * (44 g CO2 / 1 mol CO2) = 150.86 g CO2
hope this helps~ :)
To find the mass of CO2 produced, we need to use the balanced equation and the molar mass of CO2.
First, let's start by converting moles of O2 to moles of CO2 using the stoichiometry of the equation.
From the balanced equation, we see that the ratio of the moles of O2 to CO2 is 7:4.
Therefore, if 6 moles of O2 are used, the amount of CO2 produced can be calculated as follows:
(6 moles O2) x (4 moles CO2/7 moles O2) = 24/7 moles CO2
Now, to convert moles of CO2 to grams, we need to multiply by the molar mass of CO2.
The molar mass of CO2 is calculated as follows:
(1 carbon atom x 12.01 g/mol) + (2 oxygen atoms x 16.00 g/mol) = 44.01 g/mol.
Therefore, the mass of CO2 produced can be calculated as follows:
(24/7 moles CO2) x (44.01 g CO2/1 mole CO2) = 120 g CO2.
Therefore, 120 grams of CO2 will be produced.
To find the number of grams of CO2 produced, we need to use the given balanced chemical equation:
2C2H6 + 7O2 --> 4CO2 + 6H2O
First, let's calculate the moles of CO2 produced:
From the balanced equation, we can see that 2 moles of C2H6 will produce 4 moles of CO2. Therefore, the moles of CO2 produced will be:
(6 moles O2) × (4 moles CO2 / 7 moles O2) = 24/7 moles CO2
Now, we need to convert moles of CO2 to grams using the molar mass of CO2. The molar mass of CO2 is approximately 44.01 grams/mole (carbon: 12.01 g/mole, oxygen: 16.00 g/mole x 2):
(24/7 moles CO2) × (44.01 g CO2 / 1 mole CO2) = 120.06 g CO2
Therefore, approximately 120.06 grams of CO2 will be produced when 6 moles of O2 are used in the reaction.