32.9k views
2 votes
What mass of zinc is needed to react with 23.1 g of phosphoric acid according to the following equation? 3 zn + 2 h3(po4) → 3 h2 + zn3(po4)2

A) 65.4 g
B) 98.1 g
C) 130.8 g
D) 163.5 g

User Deadrunk
by
7.7k points

1 Answer

5 votes

Final answer:

To find the mass of zinc needed to react with 23.1 g of phosphoric acid, we can use a balanced chemical equation and set up a proportion.

Step-by-step explanation:

To find the mass of zinc needed to react with 23.1 g of phosphoric acid, we first need to balance the chemical equation:

3 Zn + 2 H3(PO4) → 3 H2 + Zn3(PO4)2

The balanced equation shows that 3 moles of Zn react with 2 moles of H3(PO4). We can use this information to set up a proportion:

(3 moles of Zn) / (2 moles of H3(PO4)) = (x grams of Zn) / (23.1 g of H3(PO4))

Cross-multiplying and solving for x, we get:

x = (3/2) * (23.1 g) = 34.65 g

Therefore, the mass of zinc needed to react with 23.1 g of phosphoric acid is 34.65 g.

User Igor Antonov
by
8.2k points