Question

What mass, in g, of agcl is formed from the reaction of 75.0 ml of a 0.078 m agc2h3o2 solution with 55.0 ml of 0.109 m mgcl2 solution? 2 agc2h3o2(aq) + mgcl2(aq) â 2 agcl(s) + mg(c2h3o2)2(aq)?

Answer 1

Answer:

The mass we will get from AgCl is 0.839g

Explanation:

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Let's solve this!

The reaction is:

2AgC2H3O2 + MgCl2 ---> 2AgCl + Mg (C2H3O2) 2

The data we have are:

V1 = 75mL * (1L / 1000L) = 0.075L

M1 = 0.078M

V2 = 55mL * (1L / 1000mL) = 0.055L

M2 = 0.109M

The molarity formula is

M = mol / V

We calculate the moles of each compound:

molAgC2H3O2 = 0.075L * 0.078mol / L = 0.00585mol

molMgCl = 0.055L * 0.109M = 0.005995

We calculate the compound that is in excess, and then use the limiting reagent to calculate the amount of product we will obtain.

Since there are two Cl in MgCl2, this means that we have 0.01199 mol to produce the product, but we will only use 0.00585 mol which is the amount we have of AgC2H3O2. So MgCl is the excess reagent.

Then:

AgCl mass = 0.00585mol * (2mol AgCl / 2molAgC2H3O2) * (143.34g AgCl / 1mo AgCl) = 0.839gAgCl

The mass we will get from AgCl is 0.839g

Answer 2

0.839 g of AgCl is formed.

Further explanation

Given a reaction between:

• 75.0 ml of a 0.078 M AgC₂H₃O₂ solution
• 55.0 ml of 0.109 M MgCl₂ solution

Question:

What mass, in g, of AgCl is formed?

The Process:

Step-1:

Let us find the mole numbers of both reagents:

$\boxed{ \ Concentration \ (M) = \frac{moles \ (n)}{volume \ (V)} \ } \rightarrow \boxed{ \ n = MV \ }$

AgC₂H₃O₂ → 0.078 x 0.075 = 5.85 mmol

MgCl₂ → 0.109 x 0.055 = 5.995 mmol

Step-2:

The equation for the reaction is

$\boxed{ \ 2 AgC_2H_3O_2_{(aq)} + MgCl_2_{(aq)} \rightarrow 2 AgCl_{(s)} + Mg(C_2H_3O_2)_2_{(aq)} \ }$

According to the equation, 2 mol of AgC₂H₃O₂ react with 1 mol of AgCl.

Let us check which substances will be a limiting reagent.

AgC₂H₃O₂ → $\boxed{ \ \frac{5.85}{2} = 2.925 \ }$

MgCl₂ → $\boxed{ \ \frac{5.995}{1} = 5.995 \ }$

AgC₂H₃O₂ is the limiting reagent because the test results are the smallest.

Step-3:

As AgC₂H₃O₂ is the limiting reagent, the amount of AgCl produced will be determined by the amount of AgC₂H₃O₂. Since the proportion between the mole numbers of AgC₂H₃O₂ and AgCl is one to one, their mole numbers will be equal:

$\boxed{ \ The \ mole \ of AgCl = \frac{1}{1} \times the \ mole \ of \ AgC_2H_3O_2 \ }$

So the amount of AgCl is 5.85 moles.

Step-4:

Prepare the molar mass of AgCl.

Mr = 108 + 35.5 = 143.5 g/mol

Let us find out how many mass, in g, of AgCl is formed.

$\boxed{ \ mol \ (n) = \frac{mass \ (g)}{molar \ mass \ (Mr)} \ } \rightarrow \boxed{ \ g = n \times Mr \ }$

Mass = 5.85 mmol x 143.5 g/mol

Mass = 839.475 mg ≈ 0.839 g.

Thus the amount of AgCl is formed is 0.839 g.

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