We must first write out the entire equation for this reaction which is as follows:
CO + Fe2O3 --> Fe + CO2
Now we must balance this equation which provides us with the following equation:
3 CO + Fe2O3 --> 2Fe + 3 CO2
We are told that we have excess Fe2O3, so that suggests that CO is the limiting reagent. We now simply convert the mass of Fe given to moles of Fe, and convert moles of Fe to moles of CO.
35.0 g Fe/ 55.845 g/mol = 0.627 moles Fe
0.627 moles Fe x (3 moles CO)/(2 moles Fe) = 0.940 moles CO
Now with the moles of CO present, we simply convert this back to mass using the molecular weight of CO.
0.940 moles CO x 28.01 g/mol = 26.3 g CO.
Therefore, 26.3 g of CO are needed to produce 35.0 g of Fe. Since we began with three significant figures in our starting mass, our answer must also have three significant figures.
Answer:
0.6 Molar
Explanation:
According to the Question,
In 500 mL of water ,
0.2 mol sample of MgCl is present and 0.1 mol of KCl is present.
Thus ,
Concentration of Cl anion in the solution will be the sum of Cl anion concentration from MgCl and KCl.
From MgCl , 0.2 mol of Cl anion will come and 0.1 mol of KCl anion will come ,
Thus , 0.3 mol of Cl anion will come totally .
So, Cl anion concentration will be = 0.6 M
Answer:
Keq=0.866
Explanation:
Hello,
In this case, the undergone chemical reaction is:
In such a way, since 0.0055 mol of N₂O₄ remains in the flask, one infers that the reacted amount () was:
In addition, the produced amount of NO₂ is:
Finally, considering the flask's volume, the equilibrium constant is then computed as follows:
Best regards.
Answer:
20.93 g
Explanation:
From the question given above, the following data were obtained:
Heat (Q) = 3.5 KJ
Initial temperature (T₁) = 26°C
Final temperature (T₂) = 66°C
Mass (M) =?
Next, we shall convert 3.5 KJ to J. This can be obtained as follow:
1 KJ = 1000 J
Therefore,
3.5 KJ = 3.5 KJ × 1000 J / 1 KJ
3.5 KJ = 3500 J
Next, we shall determine the change in the temperature of the water. This is illustrated:
Initial temperature (T₁) = 26°C
Final temperature (T₂) = 66°C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 66 – 26
ΔT = 40 °C
Finally, we shall determine the mass of the water. This can be obtained as follow:
Heat (Q) = 3500 J
Change in temperature (ΔT) = 40 °C
Specific heat capacity (C) = 4.18 J/gºC
Mass (M) =?
Q = MCΔT
3500 = M × 4.18 × 40
3500 = M × 167.2
Divide both side by 167.2
M = 3500 / 167.2
M = 20.93 g
Therefore, the mass of the water is 20.93 g
Answer:
0.05mole
Explanation:
C2H4+ 3O2 —> 2CO2 + 2H2O
From the equation, 1mole of ethene produced 2 moles of H2O.
Therefore, Xmol of ethene will produce 0.1mol of H2O i.e
Xmol of ethene = 0.1/2 = 0.05mole