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
When The balanced equation is:
2Al + 3CuCl2 ⇒3 Cu + 2AlCl3
So, we want to find the limiting reactant:
1- no. of moles of 2Al = MV/n = (Wt * V )/ (M.Wt*n*V) = Wt / (M.Wt *n)
where M= molarity, V= volume per liter and n = number of moles in the balanced equation.
by substitute:
∴ no. of moles of 2Al = 0.2 / (26.98 * 2)= 0.003706 moles.
2- no.of moles of 3CuCl2= M*v / n = (0.5*(15/1000)) / 3= 0.0025 moles.
So, CuCl2 is determining the no.of moles of the products.
∴The no. of moles of 3Cu = 0.0025 moles.
∴The no.of moles of Cu= 3*0.0025= 0.0075 moles.
and ∵ amount of weight (g)= no.of moles * M.Wt = 0.0075 * M.wt of Cu
= 0.0075 * 63.546 =0.477 g
Answer:
See explanation.
Explanation:
Hello!
In this case, since we know the heat of reaction per gram of reactant and we should know the total energy of reaction, but it is not there, we are going to assume it is 1200 J as usual in these problems, so you can change it to whatever your given heat is.
In such a way, we set up the math as shown below:
Which results:
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