Question

Sodium hydroxide and magnesium chloride react as shown by this equation: 2NaOH + MgCl2 → Mg(OH)2 + 2NaCl. Suppose the reaction begins with 637 milliliters of 1.35 M sodium hydroxide solution and excess magnesium hydroxide. What is the theoretical yield of magnesium hydroxide if the resulting solution has a volume of 2.82 liters? Use the periodic table and the polyatomic ion resource. The mass of magnesium hydroxide formed is grams.

Answer 1

Answer:- 25.2 g

Solution:- The balanced equation is:

$2NaOH+MgCl_2\rightarrow Mg(OH)_2+2NaCl$

From given balanced equation, there is 2:1 mol ratio between sodium hydroxide and magnesium hydroxide.

From given molarity and volume, we calculate the moles of NaOH and using mol ratio the moles of magnesium hydroxide are calculated that could easily be converted to grams on multiplying by it's molar mass.

Molar mass of $Mg(OH)_2$  = 24.305 + 2( 15.999 + 1.008)

= 24.305 + 2(17.007)

= 58.319 gram per mol

let's make the set up using dimensional analysis as:

$637mL(\frac{1L}{1000mL})(\frac{1.35molNaOH}{1L})(\frac{1molMg(OH)_2}{2molNaOH})(\frac{58.319gMg(OH)_2}{1molMg(OH)_2})$

= $25.1gMg(OH)_2$

So, 25.2 grams of magnesium hydroxide are formed.