Question

# How many grams of water are expected formed if you begin with 75.2 g of Zn(OH)2...

How many grams of water are expected formed if you begin with 75.2 g of Zn(OH)2 and 43.4 g of HCl?

Zn(OH)2 (aq) + 2 HCl(aq) → ZnCl2 (aq) + 2 H2O (l)

Number of moles of Zn(OH)2 = 75.2 g / 99.424 g/mol = 0.756 mol

Number of moles of HCl = 43.4 g / 36.46 g/mol = 1.19 mol

From the balanced equation we can say that

1 mole of Zn(OH)2 requires 2 mole of HCl so

0.756 mole of Zn(OH)2 will require

= 0.756 mole of Zn(OH)2 *(2 mole of HCl / 1 mole of Zn(OH)2)

= 1.51 mole of HCl

But we have 1.19 mole of HCl which is in short so HCl is limiting reactant

From the balanced equation we can say that

2 mole of HCl produces 2 mole of H2O so

so 1.19 mole of HCl will produce

= 1.19 mole of HCl *(2 mole of H2O / 2 mole of HCl )

= 1.19 mole of H2O

mass of 1 mole of H2O = 18.015 g

so the mass of 1.19 mole of H2O = 21.4 g

Therefore, the mass of H2O produced would be 21.4 g

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