Respuesta :
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,
[tex]\text {number of moles}=\frac{\text {mass}}{\text {molar weight}}[/tex]
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 [tex]=\frac{23.1}{98}=0.2357 \text { moles }[/tex]
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,
[tex]\text { X moles of } \mathrm{Zn}=\frac{(0.2357 \times 3)}{2}=\frac{0.707}{2}=0.3535 \text { moles of } \mathrm{Zn}[/tex]
Now, calculate mass of zinc as follows, we know
molar mass of zinc is 65 g/mol
[tex]mass\ of\ zinc = number\ of\ moles \times molar\ weight =0.3535 \times 65[/tex]
Mass of zinc = 22.98 g
