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How many moles of ZnCO3 and C6H8O7 are required to produce 30.0 mol of...

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tdj4325 | Salutatorian

Posted June 19, 2012 at 1:48 AM via web

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How many moles of ZnCO3 and C6H8O7 are required to produce 30.0 mol of Zn3(C6H5O7)2?

Here's the formula: 3 ZnCO3 (s) + 2 C6H8O7 (aq) → Zn3(C6H5O7)2 (aq) + 3 H2O (l) + 3 CO2 (g)

How many moles of ZnCO3 and C6H8O7 are required to produce 30.0 mol of Zn3(C6H5O7)2? Show your work, including all conversion factors and all units.

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thilina-g | College Teacher | (Level 1) Educator

Posted June 19, 2012 at 3:12 AM (Answer #1)

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`3 ZnCO_3(s) + 2 C_6H_8O_(7(aq))->`

`Zn_3(C_6H_5O_7)_(2(aq)) + 3 H_2O_(l) + 3 CO2_(g)`

 

According to stoichiometry of the equation, to produce 1 mol of Zn3(C6H5O7)2 you need 3 mols of ZnCO3 and 2 mols of C6H8O7.

Therefore to produce 30 mols of Zn3(C6H5O7)2

ZnCO3 mols required = 3 x 30 = 90 mol

C6H8O7 mols required = 2 x 30 = 60 mol

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