1.
If 15.6 grams of iron (III) oxide reacts with 12.8 grams of carbon monoxide to produce 9.58 g of pure iron, what are the theoretical yield and percent yield of this reaction? Be sure to show the work that you did to solve this problem.
unbalanced equation: Fe2O3 + CO Fe + CO2
#2.
How many milliliters of sodium metal, with a density of 0.97 g/mL, would be needed to produce 34.5 grams of sodium hydroxide in the single replacement reaction below? Be sure to show the work that you did to solve this problem.
unbalanced equation: Na + H2O NaOH + H2
Any help would be greatly appreciated :)
If 15.6 grams of iron (III) oxide reacts with 12.8 grams of carbon monoxide to produce 9.58 g of pure iron, what are the theoretical yield and percent yield of this reaction? Be sure to show the work that you did to solve this problem.
unbalanced equation: Fe2O3 + CO Fe + CO2
#2.
How many milliliters of sodium metal, with a density of 0.97 g/mL, would be needed to produce 34.5 grams of sodium hydroxide in the single replacement reaction below? Be sure to show the work that you did to solve this problem.
unbalanced equation: Na + H2O NaOH + H2
Any help would be greatly appreciated :)
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first, balance that equation; you need three CO molecules, and you'd make two iron atoms, and three CO2 molecules.
15.6 grams of iron oxide corresponds to 0.0975 moles
which would need three times as many moles of CO ; that is 0.2925 moles
The 12.8 grams of CO corresponds to 0.457 moles, so the CO is present in excess.
If the CO is in excess, the iron oxide must be the limiting reagent.
which means that 0.0975 moles could produce 0.195 moles of iron
and 0.195 moles of iron weighs 10.92 grams.
but you only got 9.58 g, which is 87.7% of the theoretical yield.
As for part 2: 34.5 g of NaOH is equal to 0.8625 moles
we balance the equation by having half a molecule of H2;
still leaves a ratio of 1:1 between the NaOH and the Na metal.
so we'll need 0.8625 moles of sodium metal, which is 19.8375 grams
mass/density = volume, so our volume of sodium would be 20.45 mL
15.6 grams of iron oxide corresponds to 0.0975 moles
which would need three times as many moles of CO ; that is 0.2925 moles
The 12.8 grams of CO corresponds to 0.457 moles, so the CO is present in excess.
If the CO is in excess, the iron oxide must be the limiting reagent.
which means that 0.0975 moles could produce 0.195 moles of iron
and 0.195 moles of iron weighs 10.92 grams.
but you only got 9.58 g, which is 87.7% of the theoretical yield.
As for part 2: 34.5 g of NaOH is equal to 0.8625 moles
we balance the equation by having half a molecule of H2;
still leaves a ratio of 1:1 between the NaOH and the Na metal.
so we'll need 0.8625 moles of sodium metal, which is 19.8375 grams
mass/density = volume, so our volume of sodium would be 20.45 mL
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1. balanced equation Fe2O3 + 3CO ---> 2Fe + 3CO2
first, find the limiting reactant using stoichiometry:
15.6g Fe2O3 x 1mol Fe2O3/159.69g x 3mol CO/1mol Fe2O3 x 28g CO/1mol CO= 8.2g CO required
12.8g CO x 1mol CO/28g x 1mol Fe2O3/3mol CO x 159.69g Fe2O3/1mol= 24.33g Fe2O3 required
** Fe2O3 is limiting because you need more than you have **
Then, the theoretical yield of iron in this equation is found using the mass of Fe2O3 you have, and converting it into the mass of iron that would be created using this mass:
first, find the limiting reactant using stoichiometry:
15.6g Fe2O3 x 1mol Fe2O3/159.69g x 3mol CO/1mol Fe2O3 x 28g CO/1mol CO= 8.2g CO required
12.8g CO x 1mol CO/28g x 1mol Fe2O3/3mol CO x 159.69g Fe2O3/1mol= 24.33g Fe2O3 required
** Fe2O3 is limiting because you need more than you have **
Then, the theoretical yield of iron in this equation is found using the mass of Fe2O3 you have, and converting it into the mass of iron that would be created using this mass:
12
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