Question

Approximately 100 g of PbCl4 reacted excess NH4Cl according to the following equation: PbCl4 + 2 NH4Cl → (NH4)2PbCl6. If the yield was 87.0%, how many grams of product were obtained?

please include calculations

Answer 1

The molar masses are:

PbCl₄ - 349.01 g/mol;

(NH₄)₂PbCl₆ - 455.99 g/mol.

$m \ (NH_4)_2PbCl_6=100\ g\ PbCl_4\times\frac{1\ mol\ PbCl_4}{349.01\ g\ PbCl_4}\times\frac{1\ mol\ (NH_4)_2PbCl_6}{1\ mol\ PbCl_4}\times\frac{455.99\ g\ (NH_4)_2PbCl_6}{1\ mol\ (NH_4)_2PbCl_6}\times0.870=114\ g$

Answer: 114 g