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vodomira [7]
3 years ago
5

write a balanced chemical equation for the reaction of zncl2 with excess NaOH to produce Na2Zn(oh)4 sodium zincate. what mass of

sodium zincate can be produced from 2.00 g of ZNCl2 with excess Naoh by this reaction?
Chemistry
1 answer:
timurjin [86]3 years ago
7 0
4NaOH _{(aq)}   +  ZnCl_{2}_{(aq)}   ----\ \textgreater \   Na_{2}Zn(OH)_{4}_{(ppt)}   +  2NaCl (aq)
 
mol =  \frac{mass}{molar mass}<span>
</span>∴  mol of ZnCl_{2} =  \frac{2.00 g}{[(65)+(35.5 * 2)g/mol}
                                    =  \frac{2.00 g}{136 g/mol}
                                    =  0.0147 mol

Moles of Na_{2}Zn(OH)_{4}  :
ratio of ZnCl_{2}  :  Na_{2}Zn(OH)_{4}
                       1      :       1
∴ Moles of Na_{2}Zn(OH)_{4} = 0.0147 mol

Mass = Molar Mass * Mol
∴ Mass of Na_{2}Zn(OH)_{4} = [(23 * 2)+(65)+(16 * 4) + (1*4)] g/mol * 0.0147 mol
                                                       = 179 g/mol * 0.0147 mol
                                                       = 2.63 g
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