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Given the following reaction:
Mg(OH), + 2HCI — MgCl2 + 2H2O
How many grams of MigCl, will be produced from 16.0 g of Mg
d 11.0 g of HCI?
grams (round to three significant figures)
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Answer 1

Mass = 14.3 g

Step-by-step explanation:

Given data:

Mass of Mg(OH)₂ = 16.0 g

Mass of HCl = 11.0 g

Mass of MgCl₂ = ?

Solution:

Chemical equation:

Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O

Number of moles of Mg(OH)₂ :

Number of moles = mass/ molar mass

Number of moles = 16.0 g/ 58.3 g/mol

Number of moles = 0.274 mol

Number of moles of HCl :

Number of moles = mass/ molar mass

Number of moles = 11.0 g/ 36.5 g/mol

Number of moles = 0.301 mol

Now we will compare the moles of Mg(OH)₂ and HCl with MgCl₂.

Mg(OH)₂ : MgCl₂

1 : 1

0.274 : 0.274

HCl : MgCl₂

2 : 1

0.301 : 1/2×0.301 = 0.150

The number of moles of MgCl₂ produced by HCl are less so it will limiting reactant.

Mass of MgCl₂:

Mass = number of moles × molar mass

Mass = 0.150 × 95 g/mol

Mass = 14.3 g