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...

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.

Asked on by tdj4325

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

Posted on

`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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