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

The answer to your question is LiOH and Li₂SO₄

Explanation:

In a neutralization reaction, the reactants are an acid and a base, and the products are salt and water.

H₂SO₄ is an acid, so in the options, we must look for a base and a salt.

The first option is incorrect because the compounds are written incorrectly.

The second option is correct because LiOH is a base and LI₂SO₄ is a salt and these compounds are written correctly.

The third and fourth options are wrong.