Question

Redo the experiment by clicking on Reset Experiment. Add Ca(NO3)2 to 40 g of Na2CO3 and determine at what point the masses of the two reactants react "evenly." That is, how many grams of Ca(NO3)2 must be added to just consume the 40 g Na2CO3 initially available?

Answer 1

Answer : The mass of $Ca(NO_3)_2$  added must be, 61.9 grams

Explanation : Given,

Mass of $Na_2CO_3$ = 40 g

Molar mass of $Na_2CO_3$ = 105.9 g/mol

First we have to calculate the moles of $Na_2CO_3$.

$\text{ Moles of }Na_2CO_3=\frac{\text{ Mass of }Na_2CO_3}{\text{ Molar mass of }Na_2CO_3}=\frac{40g}{105.9g/mole}=0.378moles$

Now we have to calculate the moles of $MgO$

The balance chemical reaction will be:

$Ca(NO_3)_2+Na_2CO_3\rightarrow CaCO_3+2NaNO_3$

From the balanced reaction we conclude that

As, 1 mole of $Na_2CO_3$ react with 1 mole of $Ca(NO_3)_2$

So, 0.378 mole of $Na_2CO_3$ react with 0.378 mole of $Ca(NO_3)_2$

Now we have to calculate the mass of $Ca(NO_3)_2$

$\text{ Mass of }Ca(NO_3)_2=\text{ Moles of }Ca(NO_3)_2\times \text{ Molar mass of }Ca(NO_3)_2$

Molar mass of $Ca(NO_3)_2$ = 164 g/mol

$\text{ Mass of }Ca(NO_3)_2=(0.378moles)\times (164g/mole)=61.9g$

Thus, the mass of $Ca(NO_3)_2$  added must be, 61.9 grams