Question

\[ 3 \mathrm{MnC}_{2} \mathrm{O}_{4} \cdot \mathrm{YH}_{2} \mathrm{O}(s)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{Mn}_{3} \mathrm{O}_{4}(s)+6 \mathrm{CO}_{2}(g)+\mathrm{ZH}_{2} \mathrm{O}(g) \] 1.0000 g of \( \mathrm{MnC}_{2} \mathrm{O}_{4} \cdot \mathrm{Y} \mathrm{H}_{2} \mathrm{O} \) were reacted with excess oxygen. Answer the following questions. Remember to use the appropriate number of significant figures and only round at the end of your calculations. a. Pyrolysis resulted in 0.3871 g of \( \mathrm{Mn}_{3} \mathrm{O}_{4} \). The molar mass of \( \mathrm{Mn}_{3} \mathrm{O}_{4} \) is \( 228.812 \mathrm{~g} / \mathrm{mol} \). How many moles of Mn atoms were produced from this reaction? \( \square \) moles b. All Mn atoms in your \( \mathrm{Mn}_{3} \mathrm{O}_{4} \) came from your \( \mathrm{MnC}_{2} \mathrm{O}_{4} \cdot \mathrm{YH}_{2} \mathrm{O} \) product in Part 1. How many grams of \( \mathrm{MnC}_{2} \mathrm{O}_{4} \) (the Mn oxalate hydrate if \( \mathrm{Y}=0 \) ) would be required to supply enough Mn atoms for your answer above? \( \square \) g c. How many moles of \( \mathrm{MnC}_{2} \mathrm{O}_{4} \) does this represent? \( \square \) moles d. You began your pyrolysis with 1.000 g of your \( \mathrm{MnC}_{2} \mathrm{O}_{4} \cdot \mathrm{Y} \mathrm{H}_{2} \mathrm{O} \) product. Subtract your answer to question b above from 1.000 g . This is the mass of \( \mathrm{H}_{2} \mathrm{O} \) in 1.000 g of your \( \mathrm{MnC}_{2} \mathrm{O}_{4} \cdot \mathrm{Y} \mathrm{H}_{2} \mathrm{O} \) product. How many moles of \( \mathrm{H}_{2} \mathrm{O} \) does this represent? \( \square \) mol e. The number of waters of hydration is the ratio moles of water and moles of \( \mathrm{MnC}_{2} \mathrm{O}_{4} \) in the hydrate. In other words, \( \mathrm{Y}=\mathrm{mol} \mathrm{H}_{2} \mathrm{O} / \mathrm{mol}_{\mathrm{MnC}_{2} \mathrm{O}_{4} \text {. Use your answers to questions } \mathrm{c}} \) and \( d \) to calculate \( Y \).

          \[
3 \mathrm{MnC}_{2} \mathrm{O}_{4} \cdot \mathrm{YH}_{2} \mathrm{O}(s)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{Mn}_{3} \mathrm{O}_{4}(s)+6 \mathrm{CO}_{2}(g)+\mathrm{ZH}_{2} \mathrm{O}(g)
\]
1.0000 g of \( \mathrm{MnC}_{2} \mathrm{O}_{4} \cdot \mathrm{Y} \mathrm{H}_{2} \mathrm{O} \) were reacted with excess oxygen.

Answer the following questions. Remember to use the appropriate number of significant figures and only round at the end of your calculations.
a. Pyrolysis resulted in 0.3871 g of \( \mathrm{Mn}_{3} \mathrm{O}_{4} \). The molar mass of \( \mathrm{Mn}_{3} \mathrm{O}_{4} \) is \( 228.812 \mathrm{~g} / \mathrm{mol} \). How many moles of Mn atoms were produced from this reaction?
\( \square \) moles
b. All Mn atoms in your \( \mathrm{Mn}_{3} \mathrm{O}_{4} \) came from your \( \mathrm{MnC}_{2} \mathrm{O}_{4} \cdot \mathrm{YH}_{2} \mathrm{O} \) product in Part 1. How many grams of \( \mathrm{MnC}_{2} \mathrm{O}_{4} \) (the Mn oxalate hydrate if \( \mathrm{Y}=0 \) ) would be required to supply enough Mn atoms for your answer above?
\( \square \)
g
c. How many moles of \( \mathrm{MnC}_{2} \mathrm{O}_{4} \) does this represent?
\( \square \) moles
d. You began your pyrolysis with 1.000 g of your \( \mathrm{MnC}_{2} \mathrm{O}_{4} \cdot \mathrm{Y} \mathrm{H}_{2} \mathrm{O} \) product. Subtract your answer to question b above from 1.000 g . This is the mass of \( \mathrm{H}_{2} \mathrm{O} \) in 1.000 g of your \( \mathrm{MnC}_{2} \mathrm{O}_{4} \cdot \mathrm{Y} \mathrm{H}_{2} \mathrm{O} \) product. How many moles of \( \mathrm{H}_{2} \mathrm{O} \) does this represent?
\( \square \) mol
e. The number of waters of hydration is the ratio moles of water and moles of \( \mathrm{MnC}_{2} \mathrm{O}_{4} \) in the hydrate. In other words, \( \mathrm{Y}=\mathrm{mol} \mathrm{H}_{2} \mathrm{O} / \mathrm{mol}_{\mathrm{MnC}_{2} \mathrm{O}_{4} \text {. Use your answers to questions } \mathrm{c}} \) and \( d \) to calculate \( Y \).

3 MnC2O4·YH2O(s)+2 O2(g) →Mn3O4(s)+6 CO2(g)+ZH2O(g)

1.0000 g of MnC2O4·YH2O were reacted with excess oxygen.

Answer the following questions. Remember to use the appropriate number of significant figures and only round at the end of your calculations.
a. Pyrolysis resulted in 0.3871 g of Mn3O4. The molar mass of Mn3O4 is 228.812 g / mol. How many moles of Mn atoms were produced from this reaction?
□ moles
b. All Mn atoms in your Mn3O4 came from your MnC2O4·YH2O product in Part 1. How many grams of MnC2O4 (the Mn oxalate hydrate if Y=0 ) would be required to supply enough Mn atoms for your answer above?
□
g
c. How many moles of MnC2O4 does this represent?
□ moles
d. You began your pyrolysis with 1.000 g of your MnC2O4·YH2O product. Subtract your answer to question b above from 1.000 g . This is the mass of H2O in 1.000 g of your MnC2O4·YH2O product. How many moles of H2O does this represent?
□ mol
e. The number of waters of hydration is the ratio moles of water and moles of MnC2O4 in the hydrate. In other words, Y=molH2O / molMnC2O4. Use your answers to questions c and d to calculate Y.

Added by John D.

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