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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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Mass = 14.3 g

Explanation:

Given dа ta:

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