Answer:
0.18× 10²³ molecules
Explanation:
Given data:
Mass of copper hydroxide = 3.30 g
Number of molecules = ?
Solution:
Number of moles = mass/molar mass
Number of moles = 3.30 g/97.56 g/mol
Number of moles = 0.03 mol
Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ molecules
0.03 mol × 6.022 × 10²³ molecules / 1mol
0.18× 10²³ molecules
its 40
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The activity series goes top to bottom, most active to least active elements, going: Li, K, Ba, Sr, Ca, Na, Mg, Mn, Zn, Fe, Cd, Co, Ni, Sn, Pb, H, Cu, Ag, Hg, Au.
Thus, your list of metals would go from most reactive to least reactive: Li, K, Mg, Zn, Fe, Cu, Au
Taking into account the reaction stoichiometry, 16.611 grams of Na₂CO₃ are necessary to completely react with 17.3 g of CuCl₂.
<h3>Reaction stoichiometry</h3>
In first place, the balanced reaction is:
Na₂CO₃ + CuCl₂ → CuCO₃ + 2 NaCl
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Na₂CO₃: 1 mole
- CuCl₂: 1 mole
- CuCO₃: 1 mole
- NaCl: 2 moles
The molar mass of the compounds is:
- Na₂CO₃: 129 g/mole
- CuCl₂: 134.45 g/mole
- CuCO₃: 123.55 g/mole
- NaCl: 58.45 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Na₂CO₃: 1 mole ×129 g/mole= 129 grams
- CuCl₂: 1 mole ×134.45 g/mole= 134.45 grams
- CuCO₃: 1 mole ×123.55 g/mole= 123.55 grams
- NaCl: 2 mole ×58.45 g/mole=116.9 grams
<h3>Mass of CuCl₂ required</h3>
The following rule of three can be applied: If by reaction stoichiometry 134.35 grams of CuCl₂ react with 129 grams of Na₂CO₃, 17.3 grams of CuCl₂ react with how much mass of Na₂CO₃?
mass of Na₂CO₃= (17.3 grams of CuCl₂× 129 grams of Na₂CO₃)÷ 134.35 grams of CuCl₂
<u><em>mass of Na₂CO₃= 16.611 grams</em></u>
Finally, 16.611 grams of Na₂CO₃ is required.
Learn more about the reaction stoichiometry:
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