Answer:
mass of CO = 210.42 g
mass in three significant figures = 210. g
Explanation:
Given data:
mass of Fe2O3 = 0.400 Kg
mass of CO= ?
Solution:
chemical equation:
Fe2O3 + 3CO → 2Fe + 3CO2
Now we will calculate the molar mass of Fe2O3 and CO.
Molar mass of Fe2O3 = (55.845 × 2) + (16 × 3) = 159.69 g/mol
Molar mass of CO = 12+ 16 = 28 g/mol
now we will convert the kg of Fe2O3 in g.
mass of Fe2O3 = 0.400 kg × 1000 = 400 g
number of moles of Fe2O3 = 400 g/ 159.69 g/mol = 2.505 mol
mass of CO = moles of Fe2O3 × 3( molar mass of CO)
mass of CO = 2.505 mol × 84 g/mol
mass of CO = 210.42 g
mass in three significant figures = 210. g
Answer:
The number of grams of CO that can react with 0.400 kg of [tex]Fe_2 O_3[/tex] is 210 g.
Explanation:
The balanced chemical equation is
[tex]$F e_{2} \boldsymbol{O}_{3}(s)+3 \boldsymbol{C} \boldsymbol{O}(\boldsymbol{g})>2 \boldsymbol{F e}(\boldsymbol{s})+3 \boldsymbol{C} \boldsymbol{O}_{2}(\boldsymbol{g})$[/tex]
[tex]$0.400 \mathrm{kg} \mathrm{Fe}_{2} \mathrm{O}_{3} \times \frac{1000 \mathrm{g} \mathrm{Fe}_{2} \mathrm{O}_{3}}{1 \mathrm{Kg} \mathrm{Fe}_{2} \mathrm{O}_{3}} \times \frac{1 \mathrm{mol} \mathrm{Fe}_{2} \mathrm{O}_{3}}{159.7 \mathrm{gFe}_{2} \mathrm{o}_{3}} \times \frac{3 \mathrm{molCO}}{1 \mathrm{molFe}_{2} \mathrm{o}_{3}} \times \frac{28 \mathrm{g} \mathrm{CO}}{1 \mathrm{mol} \mathrm{CO}}$[/tex]
=210g CO is the Answer
The number of grams of CO that can react with 0.400 kg of [tex]Fe_2 O_3[/tex] is 210 g.
