When 25.2g of lead (Uc) nítrate reacts with sodium chloride, what’s is the weight

19.3 g

Explanation:

Step 1: Write the balanced reaction

Pb(NO₃)₄ + 4 NaCl ⇒ 4 NaNO₃ + PbCl₄

Step 2: Calculate the moles corresponding to 25.2 g of Pb(NO₃)₄

The molar mass of 455.22 g/mol.

25.2 g × 1 mol/455.22 g = 0.0554 mol

Step 3: Calculate the moles of PbCl₄ produced from 0.0554 moles of Pb(NO₃)₄

The molar ratio of Pb(NO₃)₄ to PbCl₄ is 1:1. The moles of PbCl₄ produced are 1/1 × 0.0554 mol = 0.0554 mol.

Step 4: Calculate the mass corresponding to 0.0554 moles of PbCl₄

The molar mass of PbCl₄ is 349.01 g/mol.

0.0554 mol × 349.01 g/mol = 19.3 g