Question

How many grams of magnesium sulfate would need to be reacted with barium chloride to produce 350 grams of magnesium chloride? (2 SF, No units)

Answer 1

Illustration of the reaction is as follows :

$\sf MgSO_4 + BaCl_2 \longrightarrow MgCl_2 + BaSO_4$

• The above equation is balanced.

Mass of Magnesium Sulphate : 24 + 32 + 4 × 16 = 120g

Mass of Magnesium Chloride : 24 + 2 × 35.5 = 95g

Here, 120g of Magnesium Sulphate reacts with Barium Chloride to produce 95g of Magnesium Chloride.

Implies, 120/95g of Magnesium Sulphate would react with Barium Chloride to produce 1g of Magnesium Chloride.

Therefore, to produce 350g of Magnesium Chloride, 120/95 × 350g of Magnesium Sulphate will be required.

$\sf \: Required \: Mass \ of \ MgSO_4 = \dfrac{120}{95} \times 350 \\ \\ \longrightarrow \sf \: Required \: Mass \ of \ MgSO_4 = 442g \: (approx)$

Answer 2

• 442.105 grams of magnesium sulfate is needed to be reacted with barium chloride to produce 350 grams of magnesium chloride.

Explanation:

The reaction of given data:-

${\boxed{\red{MgSO_4+BaCl_2→MgCl_2+BaSO_4}}}$

First of all we will find out the mass of Magnesium sulphate and Magnesium Chloride.

• Magnesium Sulphate :- 24 + 32 + 4 × 16 = 120g

• Magnesium Chloride :- 24 + 2 × 35.5 = 95g

Mass of Magnesium Sulphate required to produce 350g of Magnesium Chloride

$=\dfrac{Mass\:of\: Magnesium\: Sulphate}{Mass\:of\: Magnesium\: Chloride}×350$

$=\dfrac{120g}{95g}×350$

$=\dfrac{120g}{95g}×350$

$=\dfrac{42,000g}{95g}$

$=442.105g$

Hence, 442.105g of magnesium sulfate needed to be reacted with barium chloride to produce 350 grams of magnesium chloride.

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