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

To react "evenly" with 40 g of Na2CO3, 61.89 grams of Ca(NO3)2 is required based on the stoichiometry of the reaction between Na2CO3 and Ca(NO3)2.

Step-by-step explanation:

To find out how many grams of Ca(NO3)2 are needed to react "evenly" with 40 g of Na2CO3, we need to consider the stoichiometry of the chemical reaction, which is not provided in the question but generally has the form:

Na2CO3 + Ca(NO3)2 → CaCO3 + 2 NaNO3

The molar mass of Na2CO3 is 105.99 g/mol, and the molar mass of Ca(NO3)2 has been calculated to be 164.10 g/mol. Given 40 g of Na2CO3, we first convert this mass to moles:

40 g Na2CO3 × (1 mol Na2CO3 / 105.99 g Na2CO3) = 0.377 moles Na2CO3

Now we use the stoichiometry of the reaction, which tells us that one mole of Na2CO3 reacts with one mole of Ca(NO3)2, to find the mass of Ca(NO3)2 needed:

0.377 moles Na2CO3 × (164.10 g Ca(NO3)2 / 1 mol Ca(NO3)2) = 61.89 g of Ca(NO3)2

Therefore, 61.89 grams of Ca(NO3)2 is required to react evenly with 40 g of Na2CO3.

Answer 2

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}=(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* \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)* (164g/mole)=61.9g`

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