Answer:
91.41 g of LiClO₃.
Explanation:
We'll begin by calculating the number of mole of O₂ that occupied 33.8 L. This can be obtained as follow:
22.4 L = 1 mole of O₂
Therefore,
33.8 L = 33.8 L × 1 mole / 22.4 L
33.8 L = 1.51 mole of O₂
Next, the balanced equation for the reaction.
2LiCl + 3O₂ —> 2LiClO₃
From the balanced equation above,
3 moles of O₂ reacted to produce 2 moles of LiClO₃.
Therefore, 1.51 mole of O₂ will react to produce = (1.51 × 2)/3 = 1.01 mole of LiClO₃.
Finally, we shall determine the mass of 1.01 mole of LiClO₃. This can be obtained as follow:
Mole of LiClO₃ = 1.01 mole
Molar mass of LiClO₃ = 7 + 35.5 + (3×16)
= 7 + 35.5 + 48
= 90.5 g/mol
Mass of LiClO₃ =?
Mass = mole × molar mass
Mass of LiClO₃ = 1.01 × 90.5
Mass of LiClO₃ = 91.41 g
Thus, 91.41 g of LiClO₃ were obtained from the reaction.
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Explanation:
To find the amount of product that would be formed from two or more reactants, we need to follow the following steps;
- Find the number of moles of the given reactants.
- Then proceed to determine the limiting reactant. The limiting reactant is the one in short supply which determines the extent of the reaction.
- Use the number of moles of the limiting reactant to find the number of moles of the product.
- Then use this number of moles to find the mass of the product
Useful expression:
Mass = number of moles x molar mass
They are the basic components of life or the basic building blocks of life
Do not ionize in solutions
Poor conductors of electricity/heat
Low melting/boiling points
gases or liquids at room temperature
