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Svetllana [295]
3 years ago
14

if you mixed 72.9g hydrochloric acid(aq) with 150g silver acetate(aq), what would be the limiting reagent?​

Chemistry
1 answer:
horsena [70]3 years ago
8 0

Answer:

                      Silver Acetate would be the Limiting Reagent.

Explanation:

                    The balance chemical equation for the given double displacement reaction is as;

                            HCl + AgC₂H₃O₂ → AgCl + HC₂H₃O₂

Step 1: <u>Calculate Moles of Starting Materials:</u>

Moles of HCl:

                      Moles  =  Mass / M.Mass

                      Moles  =  72.9 g / 36.46

                      Moles =  1.99 moles

Moles of AgC₂H₃O₂:

                      Moles  =  150 g / 166.91 g/mol

                      Moles  =  0.898 moles

Step 2: <u>Find out Limiting reagent as:</u>

According to balance chemical equation.

              1 mole of HCl reacts with  =  1 mole of AgC₂H₃O₂

So,

         1.99 moles of HCl will react with  =  X moles of AgC₂H₃O₂

Solving for X,

                     X =  1.99 mol × 1 mol / 1 mol

                     X =  1.99 mol of AgC₂H₃O₂

Hence, to completely consume 1.99 moles of Hydrochloric acid we will require 1.99 moles of Silver Acetate, But, we are provided with only 0.898 moles of Silver Acetate. This means Silver Acetate will consume first in the reaction therefore, it is the LIMITING REAGENT.

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Answer: Balanced molecular equation :

2Na_3PO_4(aq)+3(CH_3COO)_2Zn(aq)\rightarrow 6CH_3COONa(aq)+Zn_3(PO_4)_2(s)

Total ionic equation:

6Na^+(aq)+3PO_4^{2-}(aq)+6CH_3COO^-(aq)+3Zn^{2+}(aq)\rightarrow 6CH_3COO^-(aq)+6Na^+(aq)+Zn_3(PO_4)_2(s)  

The net ionic equation:

2PO_4^{3-}(aq)+3Zn^{2+}(aq)\rightarrow Zn_3(PO_4)_2(s)

Explanation:

Complete ionic equation : In complete ionic equation, all the substances that are strong electrolyte are present in an aqueous state as ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

When sodium phosphate and zinc acetate then it gives zinc phosphate and sodium acetate as  product.

The balanced molecular equation will be,

2Na_3PO_4(aq)+3(CH_3COO)_2Zn(aq)\rightarrow 6CH_3COONa(aq)+Zn_3(PO_4)_2(s)

The total ionic equation in separated aqueous solution will be,

6Na^+(aq)+2PO_4^{3-}(aq)+6CH_3COO^-(aq)+3Zn^{2+}(aq)\rightarrow 6CH_3COO^-(aq)+6Na^+(aq)+Zn_3(PO_4)_2(s)

In this equation, and  are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

2PO_4^{3-}(aq)+3Zn^{2+}(aq)\rightarrow Zn_3(PO_4)_2(s)

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