Magnesium reacts with iron(iii) chloride to form magnesium chloride (which can be used in fireproofing wood and in disinfectants) and iron. 3mg(s) + 2fecl3(s) → 3mgcl2(s) + 2fe(s) a mixture of 41.0 g of magnesium ( = 24.31 g/mol) and 175 g of iron(iii) chloride ( = 162.2 g/mol) is allowed to react. what mass of magnesium chloride = 95.21 g/mol) is formed?

154.0831 g

Explanation:

The formula for the calculation of moles is shown below:

Given: For

Given mass = 41.0 g

Molar mass of = 24.31 g/mol

Moles of = 41.0 g / 24.31 g/mol = 1.6865 moles

Given: For

Given mass = 175 g

Molar mass of = 162.2 g/mol

Moles of = 175 g / 162.2 g/mol = 1.0789 moles

According to the given reaction:

3 moles of Mg react with 2 moles of

1 mole of Mg react with 2/3 moles of

1.6865 mole of Mg react with (2/3)*1.6865 moles of

Moles of = 1.1243 moles

Available moles of = 1.0789 moles

Limiting reagent is the one which is present in small amount. Thus, is limiting reagent. (1.0789 < 1.1243)

The formation of the product is governed by the limiting reagent. So,

2 moles of gives 3 moles of magnesium chloride

1 mole of gives 3/2 moles of magnesium chloride

1.0789 mole of gives (3/2)*1.0789 moles of magnesium chloride

Moles of magnesium chloride = 1.61835 moles

Molar mass of magnesium chloride = 95.21 g/mol

Mass of magnesium chloride = Moles × Molar mass = 1.61835 × 95.21 g = 154.0831 g

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer

b.

the accuracy of the data.