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For the following chemical reaction, given 2.5 gram MgSO4 with an excess of Ba(NO3)2, calculate the theoretical mass of BaSO4 precipitate produced from the reaction (in grams). The following conceptual plan is designed to help you convert the mass of the limiting reactant (MgSO4) to the mass of the desired product (BaSO4).

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Answer:

The answer to your question is: 4.9 g of BaSO4

Step-by-step explanation:

Data

MgSO4 = 2.5 g

Ba(NO3)2 = excess reactant

BaSO4 = ?

Reaction

MgSO4 + Ba(NO3)2 ⇒ BaSO4 + Mg(NO3)2

MW MgSO4 = 24 + 32 + 64

= 120 g

MW BaSO4 = 137 + 32 + 64

= 233 g

120 g of MgSO4 ------------------ 233g of BaSO4

2.5 g ------------------ x

x = 4.9 g of BaSO4

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