How many liters of a 3.0M H3(PO4) solution are required to react with 4.5 g of zinc?

Question

Owen Ambrose · Accepted Answer

#0.015"L H"_3("PO"_4)# Write a balanced equation #3Zn(s)+2H_3(PO_4)(aq)->Zn_3(PO_4)_2(s)+ 3H_2(g)# Stoichiometry #4.5 cancel("g Zn")* (1cancel("mol Zn"))/ (65.38 cancel("g Zn")) * (2 cancel("mol H"_3("PO"_4) ))/ (3 cancel("mol Zn"))* (1 "L H"_3("PO"_4))/ (3 cancel("mol H"_3("PO"_4)))=# #0.015"L H"_3("PO"_4)#

Owen Ambrose · Answer

undefined To determine the volume of a 3.0 M H3PO4 solution needed to react with 4.5 g of zinc, we use stoichiometry.

Step 1: Write the balanced chemical equation:
2 H3PO4 + 3 Zn -> 3 H2 + Zn3(PO4)2

Step 2: Calculate the molar mass of zinc (Zn):
Molar mass of Zn = 65.38 g/mol

Step 3: Calculate the number of moles of zinc:
Number of moles = Mass / Molar mass
Number of moles = 4.5 g / 65.38 g/mol

Step 4: Determine the stoichiometric ratio between H3PO4 and Zn:
From the balanced equation, 3 moles of Zn react with 2 moles of H3PO4.

Step 5: Calculate the number of moles of H3PO4 needed:
Number of moles of H3PO4 = (Number of moles of Zn) * (2 moles H3PO4 / 3 moles Zn)

Step 6: Calculate the volume of H3PO4 solution using its molarity:
Volume = (Number of moles of H3PO4) / (Molarity)
Volume = (Number of moles of H3PO4) / (3.0 mol/L)

Step 7: Substitute the calculated values:
Volume = (4.5 g / 65.38 g/mol) * (2 mol / 3 mol) / (3.0 mol/L)

Step 8: Calculate the volume:
Volume ≈ 0.043 L or 43 mL

Approximately 43 milliliters of the 3.0 M H3PO4 solution are required to react with 4.5 grams of zinc.