Question

I need help with this problem and when solving it can you go in depth on how to balance this equation.

Answer 1

Firstly, the equation needs to be balanced.

Let's start by balancing Na, which we can do by adding the coefficient 2 to NaNO₃

$Na_2S+Cu(NO_3)_2\to2NaNO_3_{}+CuS$

Now, we can check for the other elements.

Cu and S are balanced and the group NO3 is also balanced, so N and O are balanced.

Now, with the balanced equation, we need to consult the molar masses of NaNO₃ and Na₂S, which can be calculated from the molar masses of the elements on them:

$\begin{gathered} M_{NaNO_{3}}\approx84.99467g\/mol \\ M_{Na_{2}S}\approx78.04454g\/mol \end{gathered}$

Now, we first convert the mass of NaNO₃ to number of moles:

$n_{NaNO_3}=\frac{m_{NaNO_3}}{M_{NaNO_{3}}}=\frac{75.19g}{84.99467\/mol}=0.884643\ldots mol$

Since the stoichiometry of Na₂S and NaNO₃ is 1 : 2, if we want 0.884643...mol of NaNO₃, we need half as much of Na₂S:

$\begin{gathered} \frac{n_{Na_2S}}{1}=\frac{n_{NaNO_3}}{2} \\ n_{Na_2S}=\frac{0.884643\ldots mol}{2}=0.442321\ldots mol \end{gathered}$

And now, we convert it to mass of Na₂S:

$m_{Na_2S}=n_{Na_2S}\cdot M_{Na_2S}=0.442321\ldots mol\cdot78.04454g\/mol=34.5208\ldots g$

Now, we need to round to 4 significant figures:

$m_{Na_{2}S}\approx34.52g$

So, the answer is 34.52g of NaNO₃.