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Leya [2.2K]
4 years ago
7

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 th

e 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?
Chemistry
1 answer:
Radda [10]4 years ago
6 0

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

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