A + 2B + 5C -> AB C + 20 When 0.5 moles of A and 1.2 moles of B were taken 0.8 moles of C) were produced. Hence the limiting reagent is
Answers
⇝ We Know :-
Mass of Magnesium (Mg) = 24 u
Mass of Sulphur (S) = 32 u
Mass of Oxygen (O) = 16 u
Mass of Hydrogen (H) = 1 u
⇝ Calculating Molar Mass of MgSO₄.7H₂O :
Molar Mass = Mass of Mg + Mass of Sulphur + 11 × Mass of Oxygen + 14 × Mass of Hydrogen.
❒ Using Above written Masses :
\begin{gathered}\text{ Molar mass} = 24 + 32 + 11 \times 16 + 14 \times 1 \\ \end{gathered} Molar mass=24+32+11×16+14×1
\begin{gathered} = 24 + 32 + 176 + 14 \\ \end{gathered}=24+32+176+14
\large\red{ \bf = 246 \: u}=246u
⇝ Calculating Molar Mass of 7H₂O :
Molar Mass = 7 × ( 2× Mass of Hydrogen + Mass of Oxygen)
❒ Using Respective Masses :
\begin{gathered}\text{ Molar mass} = 7 \times(2 \times 1 +16)\\ \end{gathered} Molar mass=7×(2×1+16)
\begin{gathered} =7\times( 18) \\ \end{gathered}=7×(18)
\large\red{ \bf = 126 \: u}=126u
⇝ Calculating Reqrd. percentage :-
\begin{gathered} \purple{\sf \small \% \: of \: water = \frac{mass \: of \: water \: in \: molecule}{mass \: of \: molecule \: } \times 100} \\ \end{gathered}%ofwater=massofmoleculemassofwaterinmolecule×100
\begin{gathered} = \frac{126}{246} \times 100 \\ \end{gathered}=246126×100
\begin{gathered} = 0.5121 \times 100 \\ \end{gathered}=0.5121×100
\bf= 51.21\%=51.21%
Hence,
\large\underline{\pink{\underline{\frak{\pmb{Required \: \% = 51.21\%}}}}}Required%=51.21%Required%=51.21%