Respuesta :
Answer:
a) Citric acid is the limiting reactant.
b) 0.45804 grams of carbon dioxide formed.
Explanation:
[tex]2NaHCO_3+H_3C_6H_5O_7\rightarrow Na_2(HC_6H_5O_7)+2 H_2O+2CO_2[/tex]
a) Moles of sodium bicarbonate :
[tex]\frac{1.00 g}{84 g/mol}=0.01190 mol[/tex]
According to reaction,2 moles sodium bicarbonate reacts with 1 mole of citric acid.
Then 0.01190 moles of sodium bicarbonate will react with:
[tex]\frac{1}{2}\times 0.01190 mol=0.005950 mol[/tex] of citric acid
Moles of citric acid :
[tex]\frac{1 g}{192.12 g/mol}=0.005205 mol[/tex]
According to reaction, 1 moles of citric acid reacts with 2 moles sodium bicarbonate.
Then 0.005205 moles of citric acid will react with:
[tex]\frac{2}{1}\times 0.005205 mol=0.01041 mol[/tex] 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:
[tex]\frac{2}{1}\times 0.005205 mol=0.01041 mol[/tex] 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.
