Answer:
This is and ADDITION REACTION
Explanation:
Because your putting a compound and an element together
Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.
Solution : Given,
Mass of Cu = 300 g
Molar mass of Cu = 63.546 g/mole
Molar mass of
= 183.511 g/mole
- First we have to calculate the moles of Cu.

The moles of Cu = 4.7209 moles
From the given chemical formula,
we conclude that the each mole of compound contain one mole of Cu.
So, The moles of Cu = Moles of
= 4.4209 moles
- Now we have to calculate the mass of
.
Mass of
= Moles of
× Molar mass of
= 4.4209 moles × 183.511 g/mole = 866.337 g
Mass of
= 866.337 g = 0.8663 Kg (1 Kg = 1000 g)
Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.
The amount of Al2O3 in moles= 1.11 moles while in grams = 113.22 grams
<em><u>calculation</u></em>
2 Al + Fe2O3 → 2Fe + Al2O3
step 1: find the moles of Al by use of <u><em>moles= mass/molar mass </em></u>formula
= 60.0/27= 2.22 moles
Step 2: use the mole ratio to determine the moles of Al2O3.
The mole ratio of Al : Al2O3 is 2: 1 therefore the moles of Al2O3= 2.22/2=1.11 moles
Step 3: finds the mass of Al2O3 by us of <u><em>mass= moles x molar mass</em></u><em> </em>formula.
The molar mass of Al2O3 = (2x27) +( 16 x3) = 102 g/mol
mass is therefore= 102 g/mol x 1.11= 113.22 grams
Answer:
<u></u>
Explanation:
<u>1. Balanced molecular equation</u>

<u>2. Mole ratio</u>

<u>3. Moles of HNO₃</u>
- Number of moles = Molarity × Volume in liters
- n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃
<u>4. Moles Ba(OH)₂</u>
- n = 0.700M × 0.0310 liter = 0.0217 mol
<u>5. Limiting reactant</u>
Actual ratio:

Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.
Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.
<u>6. Final molarity of Ba(OH)₂</u>
- Molarity = number of moles / volume in liters
- Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M