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Alex777 [14]
3 years ago
11

Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. How many grams of di

nitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely? Hint: Write a balanced chemical reaction first, then find which one is the limiting reactant.
Chemistry
1 answer:
pychu [463]3 years ago
8 0

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)

Moles of calcium nitrate = \frac{31.3 g}{164 g/mol}=0.1908 mol

Moles of ammonium fluoride = \frac{38.7 g}{37 g/mol}=1.046 mol

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 :

\frac{1}{2}\times 1.046 mol=0.523 mol 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:

\frac{2}{1}\times 0.1908 mol=0.3816 molof 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.

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