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)?

The mass we will get from AgCl is 0.839g

Explanation:

Hello!

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