Question

Hello someone please teach me how to do the last three questionda is HCI b is Na2Co3 and c is 8 moles NaCi

Answer 1

Answer:

d) 464g of NaCl.

e) 424g of the excess reactant.

f) 106g of Na2CO3 remain in excess.

Explanation:

d) From part c, we know that 8 moles of NaCl can be produced. To calculate the grams of NaCl we have to convert the 8 moles to grams, using the molar mass of NaCl:

- NaCl molar mass: 58g/mol

- Conversion:

$8moles*\frac{58g}{1mole}=464g$

So, 464g of NaCl can be produced.

e) We know from part b hat the excess reactant is Na2CO3. From the balanced reaction, we know that 2 moles of HCl react with 1 mole of Na2CO3, so with the 8 moles of the limiting reactant, we can calculate the moles of Na2CO3 that will be needed:

$\begin{gathered} 2molesHCl-1moleNa_2CO_3 \\ 8molesHCl-x=\frac{8molesHCl*1moleNa_2CO_3}{2molesHCl} \\ x=4moleNa_2CO_3 \end{gathered}$

So, only 4 moles of the excess reactant will react. We can calculate tce grams using the molar mass of Na2CO3:

- Na2CO3 molar mass: 106g/mol

- Conversion:

$4moles*\frac{106g}{1mole}=424g$

Finally, 424g of the excess reactant react.

f) To calculate the grams of the excess reactant that reman inexcess, it is necessary to convert the 5 moles to grmaams, using the molar mass of Na2CO3:

- Conversion:

$5moles*\frac{106g}{1mole}=530g$

Now we have to subtract the 530g minus the 424g that reacted of Na2CO3:

530g-424g=106g

Finaly, 106 ofNa2CO3r emain in ecxcess.