Question

Can someone please explain the last three questions to me thank you

Answer 1

d) 464g of NaCl.

e) 424g of the excess reactant.

f) 106g of Na2CO3 remain in excess.

Step-by-step 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: g/mol

- Conversion:

`8moles*(58g)/(1mole)=464g`

So, 464g of NaCl can be produced.

e) We know from part b that 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=(8molesHCl*1moleNa_2CO_3)/(2molesHCl) \\ x=4moleNa_2CO_3 \end{gathered}`

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

- Na2CO3 molar mass: 106g/mol

- Conversion:

`4moles*(106g)/(1mole)=424g`

Finally, 424g of the excess reactant react.6

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

`5moles*(106g)/(1mole)=530g`

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

530g - 424g = 106g

Finally, 106g of Na2CO3 remain in excess.