how many grams of CO are needed to react with an excess of Fe2O3 to produce 209.7gFe? Fe2O3(s) + 3CO(g) => 3CO2(g) + 2Fe(s)
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To determine the number of grams of CO needed to react with an excess of Fe2O3, we need to use stoichiometry.
First, we need to calculate the molar mass of Fe2O3 and CO.
The molar mass of Fe2O3 is calculated as follows:
2(Fe atomic mass) + 3(O atomic mass)
2(55.845 g/mol) + 3(16.00 g/mol) = 159.69 g/mol
The molar mass of CO is calculated as follows:
12.01 g/mol + 16.00 g/mol = 28.01 g/mol
Now we can set up a stoichiometric ratio based on the balanced equation:
Fe2O3(s) + 3CO(g) => 3CO2(g) + 2Fe(s)
According to the balanced equation, we know that the ratio of Fe2O3 to CO is 1:3.
Next, we need to convert the given mass of Fe to moles using the molar mass of Fe:
209.7 g / (55.845 g/mol) = 3.75 mol Fe
Since the molar ratio of Fe2O3 to CO is 1:3, we can multiply the number of moles of Fe by the stoichiometric ratio to determine the moles of CO required:
3.75 mol Fe * 3 mol CO / 2 mol Fe2O3 = 5.63 mol CO
Finally, we can convert the moles of CO to grams using the molar mass of CO:
5.63 mol CO * (28.01 g/mol) = 157.69 g CO
Therefore, approximately 157.69 grams of CO are needed to react with an excess of Fe2O3 to produce 209.7 grams of Fe.