Respuesta :
Answer:
16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant. Explanation:
[tex]Ca(NO_3)_2(aq)+2NH_4F(aq)\rightarrow CaF_2(aq)+2N_2O(g)+4H_2O(g)[/tex]
Moles of calcium nitrate = [tex]\frac{31.3 g}{164 g/mol}=0.1908 mol[/tex]
Moles of ammonium fluoride = [tex]\frac{38.7 g}{37 g/mol}=1.046 mol[/tex]
According to reaction , 2 moles of ammonium fluoride reacts with 1 mole of calcium nitrate.
Then 1.046 moles of ammonium fluoride will react with :
[tex]\frac{1}{2}\times 1.046 mol=0.523 mol[/tex] calcium nitarte .
This means that ammonium fluoride is in excess amount and calcium nitrate is in limiting amount.
Hence, calcium nitrate is a limiting reactant.
So, amount of dinitrogen monoxide will depend upon moles of calcium nitrate.
So, according to reaction , 1 mole of calcium nitarte gives 2 moles of dinitrogen monoxide gas .
Then 0.1908 moles of calcium nitrate will give:
[tex]\frac{2}{1}\times 0.1908 mol=0.3816 mol[/tex]of dinitrogen monoxide gas.
Mass of 0.03816 moles of dinitrogen monoxide gas:
0.03816 mol × 44 g/mol = 16.79 g
16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant.
