The reaction is given as
Fe2O3 (s)+ 3CO(g)--->3CO2(g)+ 2Fe(s)
No.of moles=mass in gram/molar mass
As for Fe mole =156.2g/55.847=2.7969~2.797
The ratio b/w CO and Fe is 3:2
Moles of CO needed= 2.797x3/2=4.1955
Mass of CO needed= 4.195mol x 28.01g/mol= 117.515g
Answer:
1.2×10² mmole of Na₂S₂O₃
Explanation:
From the question given above, the following data were obtained:
Volume = 0.6 L
Molarity = 0.2 mol/L
Mole of Na₂S₂O₃ =?
Molarity is simply defined as the mole of solute per unit litre of water. Mathematically, it is expressed as:
Molarity = mole /Volume
With the above formula, we can obtain the number of mole of Na₂S₂O₃ in the solution as illustrated below:
Volume = 0.6 L
Molarity = 0.2 mol/L
Mole of Na₂S₂O₃ =?
Molarity = mole /Volume
0.2 = Mole of Na₂S₂O₃ / 0.6
Cross multiply
Mole of Na₂S₂O₃ = 0.2 × 0.6
Mole of Na₂S₂O₃ = 0.12 mole
Finally, we shall convert 0.12 mole to millimole (mmol). This can be obtained as follow:
1 mole = 1000 mmol
Therefore,
0.12 mole = 0.12 mole × 1000 mmol / 1 mole
0.12 mole = 120 = 1.2×10² mmole
Thus, the chemist added 1.2×10² mmole of Na₂S₂O₃
Carbon = 12.010. Oxygen = 15.999 x 2 15.999 x 2 = 31.998 + 12.010 = 44.008 \frac{37.15 grams * 1 mole CO2}{44.008 grams}
It mean it consisted of 1 g of lead and 0.077 g of O2.
divide these numbers by molar mas.
1/82=0.012 Pb /0.004 = 3
0.077/16= 0.004 O /0.004 =1
Pb3O
