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If 1000 ml (1 L) of CH₃COOH contain 1.25 mol
let 250 ml of CH₃COOH contain x
⇒ x =
= 0.3125 mol
∴ moles of CH₃COOH in 250ml is 0.3125 mol
Now, Mass = mole × molar mass
= 0.3125 mol × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol
= 18.75 g
∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is <span>18.75 g</span>
let 250 ml of CH₃COOH contain x
⇒ x =
= 0.3125 mol
∴ moles of CH₃COOH in 250ml is 0.3125 mol
Now, Mass = mole × molar mass
= 0.3125 mol × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol
= 18.75 g
∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is <span>18.75 g</span>
Answer: The balanced equation is .
Explanation:
The given reaction equation is as follows.
Number of atoms present on reactant side are as follows.
- Li = 1
- H = 1
= 1
Number of atoms present on product side are as follows.
- Li = 1
- H = 2
To balance this equation, multiply Li by 2 and by 2 on reactant side. Also, multiply
by 2 on product side.
Hence, the equation can be rewritten as follows.
Now, number of atoms present on reactant side are as follows.
- Li = 2
- H = 2
Number of atoms present on product side are as follows.
- Li = 2
- H = 2
As there are same number of atoms on both reactant and product side. Hence, the equation is now balanced.
Thus, we can conclude that the balanced equation is .