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(Solved): molar ratios with unbalanced equation  Molar Ratios: The "alum" used in cooking is potassium al ...



molar ratios with unbalanced equation 

Molar Ratios: The alum used in cooking is potassium aluminum sulfate hydrate, \( \mathrm{KAl}\left(\mathrm{SO}_{4}\right)_{
Molar Ratios: The "alum" used in cooking is potassium aluminum sulfate hydrate, \( \mathrm{KAl}\left(\mathrm{SO}_{4}\right)_{2} \cdot \times \mathrm{H}_{2} \mathrm{O} \). The find the value of \( x_{1} \) you heated a \( 4.74 \mathrm{~g} \) sample of this hydrated compound. The compound lost \( 2.16 \mathrm{~g} \) of \( \mathrm{H}_{2} \mathrm{O} \). Find the value of \( x \). UNBALANCED equation: \( \left.\mathrm{KAl}\left(\mathrm{SO}_{4}\right)_{2} \cdot \times \mathrm{H}_{2} \mathrm{O} \mathrm{s}\right) \stackrel{\Delta}{\longrightarrow} \mathrm{KAl}\left(\mathrm{SO}_{4}\right)_{2}(\mathrm{~s})+\times \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \)


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