Answer to Question #314988 in General Chemistry for Anon

Question

What is the mass (in grams) of 1.70M sodium sulfide solution that would react with 105 g of cadmium nitrate solution?

Na2S + Cd(NO3)2 = CdS+2NaNO3

The answer is 34.66g. I'd like to know what exactly are the equations that got to that?

Answer 1

This question is a bit confusing as both masses are meant to be of pure substances, not solutions. The mass of the solution cannot be determined without knowing its density, and the result in that case would be different.

The molar masses are:

Cd(NO₃)₂ - 236.42 g/mol

Na₂S - 78.04 g/mol

According to the equation, 1 mole of Cd(NO₃)₂ reacts with 1 mole os Na₂S. Therefore, only one mass-to-mass conversion is required here:

$m(Na_2S)=105\ g\ Cd(NO_3)_2\times\frac{1\ mol\ Cd(NO_3)_2}{236.42\ g\ Cd(NO_3)_2}\times\frac{1\ mol\ Na_2S}{1\ mol\ Cd(NO_3)_2}\times\frac{78.04\ g\ Na_2S}{1\ mol\ Na_2S}=34.66\ g$

Answer: 34.66 g