Respuesta :
Answer:
Mass of Analyte H2C2O4 = 18.18 grams
Explanation:
The balanced equation is:
[tex]H_{2}C_{2}O_{4}+2KOH\rightarrow K_{2}C_{2})_{4}+H_{2}O[/tex]
The Molarity of Oxalic acid needed can be determined using:[tex]M_{1}V_{1}=M_{2}V_{2}[/tex]
Here M1 = Molarity of Oxalic acid
V1 = Volume of oxalic acid
M2 = Molarity of KOH and V2 = Volume of KOH
M1 = ? , V1 = 16.8 mL , M2 = 0.501 M , V2 = 13.54 mL
Put the values and solve for M2,
[tex]M_{1}V_{1}=M_{2}V_{2}[/tex]
[tex]2\times M_{1}(16.8)=0.501\times 13.54[/tex]
We are multiplying by 2 because , 1 H2C2O4 needs 2 KOH
[tex]M_{1}=\frac{0.501\times 13.54}{2\times 16.8}[/tex]
[tex]M_{1}=0.202M[/tex]
Hence Molarity of H2C2O4 =
0.202 M
Molar mass = H2C2O4 = 2x1 +2x12 + 4x16
= 2+24+64 = 90 gram
[tex]Mole =molarity\times Molar\ mass[/tex]
[tex]Moles=0.202\times 90[/tex]
18.18 gram
Mass of Analyte is 18.18 grams
Balanced chemical equation:
[tex]H_2C_2O_4+2KOH--->(K_2C_2)_4+ H_2O[/tex]
Dilution formula:
Dilution calculations can be performed using the formula [tex]M_1V_1 = M_2V_2[/tex]. A serial dilution is a series of stepwise dilutions, where the dilution factor is held constant at each step.
M₁ = Molarity of Oxalic acid
V₁ = Volume of oxalic acid
M₂ = Molarity of KOH and V₂ = Volume of KOH
M₁ = ? , V₁ = 16.8 mL , M₂ = 0.501 M , V₂ = 13.54 mL
Substituting values in above formula:
[tex]M_1V_1=M_2V_2\\\\2*M_1(16.8)=0.501*13.54\\\\M_1=0.202M[/tex]
Thus, Molarity of oxalic acid is 0.202M.
Molar mass of oxalic acid = 90 g/mol
[tex]\text{Mass}=\text{Molarity}*\text{Molar mass}\\\\\text{Mass}=0.202M*90g/mol\\\\\text{Mass}=18.8 g[/tex]
Thus, mass of analyte is 18.18 grams.
Find more information about Molarity here:
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