Answer:
The answer to your question is 160 g of Fe₂O₃
Explanation:
Data
mass of Fe = 112 g
mass of CO = in excess
mass of Fe₂O₃ = ?
Balanced chemical reaction
Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂
Process
1.- Calculate the molar mass of Fe₂O₃ and Fe
Molar mass Fe₂O₃ = (56 x 2) + (16 x 3) = 112 + 48 = 160 g
atomic mass of Fe = 56
2.- Use proportions to calculate the mass of Fe₂O₃ needed
160 g of Fe₂O₃ ------------------- 2(56) g of Fe
x g of Fe₂O₃ ------------------ 112 g of Fe
x = (112 x 160) / 2(56)
x = 17920/112
x = 160 g of Fe₂O₃
! mole of CO2 at STP has a volume of 22.4 liters
88 grams = 2 moles
so the required volume = 2*22.4 = 44.8 liters
Mass of Cl₂ : 164.01 g
<h3>Further explanation</h3>
A mole is a number of particles(atoms, molecules, ions) in a substance
This refers to the atomic total of the 12 gr C-12 which is equal to 6.02.10²³, so 1 mole = 6.02.10²³ particles
Can be formulated :
N = n x No
N = number of particles
n = mol
No = 6.02.10²³ = Avogadro's number
mol Cl₂ :
mass Cl₂(MW=71 g/mol) :
Answer:
142.82 g
Explanation:
The following data were obtained from the question:
Volume of water = 12 mL
Volume of water + gold = 19.4 mL
Density of gol= 19.3 g/cm³
Mass of gold =.?
Next, we shall determine the volume of the gold. This can be obtained as follow:
Volume of water = 12 mL
Volume of water + gold = 19.4 mL
Volume of gold =.?
Volume of gold = (Volume of water + gold) – (Volume of water)
Volume of gold = 19.4 – 12
Volume of gold = 7.4 mL
Finally, we shall determine the mass of the gold as follow:
Note: 1 mL is equivalent to 1 cm³
Volume of gold = 7.4 mL
Density of gol= 19.3 g/cm³ = 19.3 g/mL
Mass of gold =?
Density = mass /volume
19.3 = mass of gold /7.4
Cross multiply
Mass of gold = 19.3 × 7.4
Mass of gold = 142.82 g
Therefore, the mass of the gold pebble is 142.82 g
