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:
![Ca(NO_3)_2(aq)+2NH_4F(aq)\rightarrow CaF_2(aq)+2N_2O(g)+4H_2O(g)](https://tex.z-dn.net/?f=Ca%28NO_3%29_2%28aq%29%2B2NH_4F%28aq%29%5Crightarrow%20CaF_2%28aq%29%2B2N_2O%28g%29%2B4H_2O%28g%29)
Moles of calcium nitrate = ![\frac{31.3 g}{164 g/mol}=0.1908 mol](https://tex.z-dn.net/?f=%5Cfrac%7B31.3%20g%7D%7B164%20g%2Fmol%7D%3D0.1908%20mol)
Moles of ammonium fluoride = ![\frac{38.7 g}{37 g/mol}=1.046 mol](https://tex.z-dn.net/?f=%5Cfrac%7B38.7%20g%7D%7B37%20g%2Fmol%7D%3D1.046%20mol)
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 :
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:
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.