Question

In a certain experiment 1.00 g of sodium bicarbonate and 1.00 g of citric acid are allowed to react. (a) which is the limiting reactant? (b) how many grams of carbon dioxide form? (c) how many

Answer 1

Answer:

a) Citric acid is the limiting reactant.

b) 0.45804 grams of carbon dioxide formed.

Explanation:

$2NaHCO_3+H_3C_6H_5O_7\rightarrow Na_2(HC_6H_5O_7)+2 H_2O+2CO_2$

a) Moles of sodium bicarbonate  :

$\frac{1.00 g}{84 g/mol}=0.01190 mol$

According to reaction,2 moles sodium bicarbonate reacts with 1 mole of citric acid.

Then 0.01190 moles of sodium bicarbonate will react with:

$\frac{1}{2}\times 0.01190 mol=0.005950 mol$ of citric acid

Moles of citric acid :

$\frac{1 g}{192.12 g/mol}=0.005205 mol$

According to reaction, 1 moles of citric acid reacts with 2 moles sodium bicarbonate.

Then 0.005205 moles of citric acid will react with:

$\frac{2}{1}\times 0.005205 mol=0.01041 mol$ of sodium bicarbonate.

Citric acid is the limiting reactant.

b) Since, citric acid is in limiting amount, amount of carbon dioxide produce  will depend upon on citric acid .

According to reaction , 1 mole of citrc acid gives 2 moles of carbon dioxide.

Then 0.005205 moles of citric acid will give:

$\frac{2}{1}\times 0.005205 mol=0.01041 mol$ of carbon dioxide

Mass of 0.01041 moles of carbon dioxide:

0.01041 mol × 44 g/mol= 0.45804 g

0.45804 grams of carbon dioxide formed.

Answer 2

The complete balanced chemical reaction for this is:

C6H8O7 + 3 NaHCO3 ---> 3 H2O + 3 CO2 + Na3C6H5O7

 

First calculate for the number of moles of each reactant.

moles C6H8O7 = 1 g / (192.124 g / mol) = 5.2 * 10^-3 mol

moles NaHCO3 = 1 g / (84.01 g / mol) = 11.9 * 10^-3 mol

 

A. The ratio of the reactant from the chemical reaction is 3NaHCO3:1C6H8O7, while the given chemicals are in the ratio of:

11.9 * 10^-3NaHCO3: 5.2 * 10^-3 C6H8O7 = 2.29NaHCO3:1C6H8O7

 

Therefore this means that there is less amount of NaHCO3 supplied than what is required therefore the limiting reactant is:

NaHCO3

 

B. We calculate based on the limiting reactant.

mass CO2 = 11.9 * 10^-3 mol NaHCO3 (3 mol CO2/1 mol NaHCO3) (44.01 g/mol)

mass CO2 = 1.57 g

 

C. I believe what is asked here is the amount of excess reactant which remains. The excess reactant is C6H8O7.

mass C6H8O7 left = [5.2 * 10^-3 mol – (11.9 * 10^-3 mol NaHCO3 (1 mol C6H8O7/ 3 mol NaHCO3))] * (192.124 g / mol)

mass C6H8O7 left = 0.237 g