Question

What mass of zinc is needed to react with 23.1 g of phrosphoric acid according tot he equation?

Answer 1

Complete Question:

what mass of zinc is needed to react with 23.1 g of phosphoric acid according tot he equation?

3 Zn + 2 H₃PO₄ = 3 H₂ + Zn₃(PO₄)₂

Answer:

22.98 g, the mass of zinc (Zn) is needed to react with 23.1 g of phosphoric acid according to the given equation.

Explanation:

Given chemical reaction:

           3 Zn + 2 H₃PO₄ = 3 H₂ + Zn₃(PO₄)₂

If the substance’ molecular weight is the number of grams required per mole, the number of moles expressed as a given quantity of substances shall correspond to the grams of that substance divided by the molecular weight. Hence,

              $\text {number of moles}=\frac{\text {mass}}{\text {molar weight}}$

To find no. of moles of phosphoric acid:

Given:

mass of phosphoric acid = 23.1 g

molar weight of of phosphoric acid = 97.994 = 98 g/mole

Therefore,

no. of moles of phosphoric acid $=\frac{23.1}{98}=0.2357 \text { moles }$

Look into the given chemical reaction and divide as the following:

If 2 moles of H₃PO₄ reacts with 3 moles of Zn then 0.2357 moles of H₃PO₄ reacts with X moles of Zn. Then,

   $\text { X moles of } \mathrm{Zn}=\frac{(0.2357 \times 3)}{2}=\frac{0.707}{2}=0.3535 \text { moles of } \mathrm{Zn}$

Now, calculate mass of zinc as follows, we know

molar mass of zinc is 65 g/mol

 $mass\ of\ zinc = number\ of\ moles \times molar\ weight =0.3535 \times 65$

Mass of zinc = 22.98 g