Question

A student was asked to determine the percentage of each component in a mixture of silver nitrate, AgNO3 and magnesium hydroxide, Mg(OH)2. The mass of the sample used was 3.54 g. The student extracted AgNO3 from the mixture with water and separated the insoluble Mg(OH)2 from the solution by filtration. After evaporating the filtrate to dryness, the student recovered and dried the AgNO3, and found that it weighed 1.15 g. After drying the recovered Mg(OH)2, a mass of 2.25 g was recorded. On the basis of the mass of the sample used:
1. Calculate the % AgNO3 in the mixture,
2, Calculate the % Mg(OH)2 in the mixture.
3. Calculate the total mass of the AgNO3 and Mg(OH)2 recovered.
4. Calculate the % recovery of the components, using the total mass of the substances recovered.
5. Calculate the % error for the separation of the components of the mixture.

Answer 1

Answer:

1

$k = 32.5 \%$

2

$z =63.56 \%$

3

$m_t = 3.40 \ g$

4

$j = 96.05 \%$

5

$e = 3.954 \%$

Explanation:

From the question we are told that  

   The mass of the sample  is  $m = 3.54 \ g$

    The mass of the AgNO3 is  $m_s = 1.15 g$

   The mass of  Mg(OH)2 is  $m_n= 2.25 g$

Generally the percentage of AgNO3 in the mixture is  

          $k = \frac{1.15}{3.54} * 100$

=>        $k = 32.5 \%$

Generally the percentage of Mg(OH)2 in the mixture is  

           $z = \frac{m_n}{m} * 100$

=>         $z = \frac{2.25}{3.54} * 100$

=>           $z =63.56 \%$

Generally the total mass of   the AgNO3 and Mg(OH)2 recovered is  

      $[tex]m_t = m_s +m_n$

=>    $m_t = 1.15 +2.25$  

=>    $m_t = 3.40 \ g$

Generally the percentage recovery of the components is mathematically represented as

       $j = \frac{m_t}{m} * 100$

=>     $j = \frac{3.40}{3.54} * 100$

=>     $j = 96.05 \%$

Generally the percentage error for the separation of the components of the mixture is mathematically represented as  

      $e = \frac{m- m_t}{m} * 100$

=>    $e = \frac{3.54 - 3.40}{3.54} * 100$  

=>    $e = 3.954 \%$