A reaction vessel contains 8.90 g of CO and 8.90 g of O2. How many grams of CO2 could be produced according to the following reaction?
2CO+O2 --> 2CO2
Number of moles of CO = 8.90 g / 28.01 g/mol = 0.318 mol
Number of moles of O2 = 8.90 g / 32.0 g/mol = 0.278 mol
From the balanced equation we can say that
2 mole of CO requires 1 mole of O2 so
0.318 mole of CO will require
= 0.318 mole of CO *(1 mole of O2 / 2 mole of CO)
= 0.159 mole of O2
But we have 0.278 mole of O2 which is in excess so O2 is an excess reactant and CO is limiting reactant
From the balanced equation we can say that
2 mole of CO produces 2 mole of CO2 so
0.318 mole of CO will produce
= 0.318 mole of CO *(2 mole of CO2 / 2 mole of CO)
= 0.318 mole of CO2
mass of 1 mole of CO2 = 44.01 g
so the mass of 0.318 mole of CO2 = 14.0 g
Therefore, the mass of CO2 produced would be 14.0 g
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