Respuesta :
Answer : The mass of [tex]Ca(NO_3)_2[/tex] added must be, 61.9 grams
Explanation : Given,
Mass of [tex]Na_2CO_3[/tex] = 40 g
Molar mass of [tex]Na_2CO_3[/tex] = 105.9 g/mol
First we have to calculate the moles of [tex]Na_2CO_3[/tex].
[tex]\text{ Moles of }Na_2CO_3=\frac{\text{ Mass of }Na_2CO_3}{\text{ Molar mass of }Na_2CO_3}=\frac{40g}{105.9g/mole}=0.378moles[/tex]
Now we have to calculate the moles of [tex]MgO[/tex]
The balance chemical reaction will be:
[tex]Ca(NO_3)_2+Na_2CO_3\rightarrow CaCO_3+2NaNO_3[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]Na_2CO_3[/tex] react with 1 mole of [tex]Ca(NO_3)_2[/tex]
So, 0.378 mole of [tex]Na_2CO_3[/tex] react with 0.378 mole of [tex]Ca(NO_3)_2[/tex]
Now we have to calculate the mass of [tex]Ca(NO_3)_2[/tex]
[tex]\text{ Mass of }Ca(NO_3)_2=\text{ Moles of }Ca(NO_3)_2\times \text{ Molar mass of }Ca(NO_3)_2[/tex]
Molar mass of [tex]Ca(NO_3)_2[/tex] = 164 g/mol
[tex]\text{ Mass of }Ca(NO_3)_2=(0.378moles)\times (164g/mole)=61.9g[/tex]
Thus, the mass of [tex]Ca(NO_3)_2[/tex] added must be, 61.9 grams
