What mass in grams of sodium nitrate, nano3, is produced if 20.0g of sodium azide, nan3, in a dilute aqueous solution are reacted with excess silver nitrate, agno3, according to the following chemical equation?
nan3(aq) + agno3(aq) --> agn3(s) + nano3(aq)

1) Write the balanced equation

 
NaN3(aq) + AgNO3(aq) --> AgN3(s) + NaNO3(aq)

2) State the relevant theoretical ratio for the products and reactants in which you have the focus.

1mol NaN3 / 1mol NaNO3 or  mol NaNO3 / 1mol NaN3 

3) Transform the mass data into moles using the molar masses.

- molar mass of NaN3 = 23 g/mol + 3*14g/mol = 65 g/mol

- # of moles, n = mass / molar-mass = 20.0g / 65.0 g/mol = 0.3077 mol

4) From the theoretical ratio, get the number of moles produced. In this case the ratio is 1:1, so the number of moles of NaNO3 is also 0.3077 mol

5) Convert to mass, by multiplying by the molar mass of NaNO3

- molar mass of NaNO3: 23g/mol + 14g/mol + 3*16g/mol =  85 g/mol

- mass of NaNO3 = 0.3077 mol * 85g/mol =  26.15 g

26.15 g

I’d have to say no because without oxygen in the water the fish would all be dead. so if you want to fish for dead fish, pick the ponds with the lowest oxygen concentrations.