How do i Calculate the grams of silver chloride produced from 18.5 ml of 4.5M magnesium chloride with 99.2 ml of 1.75 M silver nitrate in a step by step?

4.21 g of AgCl

3.06 g of BaCl₂ will be needed to complete the reaction

Explanation:

The first step is to determine the reaction.

Reactants: BaCl₂ and AgNO₃

The products will be the silver chloride (AgCl) and the Ba(NO₃)₂

The reaction is: BaCl₂(aq) + 2AgNO₃(aq) → 2AgCl(s) ↓ + Ba(NO₃)₂ (aq)

We determine the silver nitrate moles: 5 g . 1mol / 169.87 g = 0.0294 moles. Now, according to stoichiometry, we know that ratio is 2:2-

2 moles of nitrate can produce 2 moles of chloride, so the 0.0294 moles of silver nitrate, will produce the same amount of chloride.

We convert the moles to mass → 143.32 g / mol . 0.0294 mol = 4.21 g of AgCl.

Now, we consider the BaCl₂.

2 moles of nitrate can react to 1 mol of barium chloride

Then, 0.0294 moles of silver nitrate will react to (0.0294 . 1) /2 = 0.0147 moles. We convert the moles to mass:

0.0147 mol . 208.23 g /1mol = 3.06 g of BaCl

Explanation:

1. 1 mole of anything contains 6.022x10^23 pieces  

d  

2. a  

3. c, number  

5. 1.24x10^24atoms / 6.022x10^23atoms/mole = 2moles h2 atoms  

6. fe2br3 mass = 2moles x mass fe + 3 moles x mass br  

molar mass = grams / number of moles = g/mole  

to find moles = g / molar mass  

to find mass = g = molar mass / moles

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8