We are given the amounts of two reactants, so this is a **limiting reactant** problem.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

**1. Gather all the information** in one place with molar masses above the formulas and everything else below the formulas.

#M_r:color(white)(mmmml) 107.87color(white)(mmmmmmmmmmmll)30.01#

#color(white)(mmmmmmm)"3Ag" + "4HNO"_3 → "3AgNO"_3 + "NO" + 2"H"_2"O""#

#"Mass/g":color(white)(mmm)216#

#"Amt/mol:"color(white)(mm)2.002color(white)(ml)1.70#

#"Divide by:"color(white)(mmm)3color(white)(mmm)4#

#"Moles rxn:"color(white)(mll)0.6675color(white)(ll)"0.425"#

#"Moles of Ag" = 216 color(red)(cancel(color(black)("g Ag"))) × ("1 mol Ag")/(107.87 color(red)(cancel(color(black)("g Ag")))) = "0.6674 mol Ag"#

#"Moles of HNO"_3 = 0.850 color(red)(cancel(color(black)("L HNO"_3))) × ("2.00 mol HNO"_3)/(1 color(red)(cancel(color(black)("L HNO"_3)))) = "1.70 mol HNO"_3#

**2. Identify the limiting reactant**

An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will give:

You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

I did that for you in the table above.

#"HNO"_3# is the limiting reactant because it gives the fewest moles of reaction.

**3. Calculate the theoretical moles of #"NO"#**

#"Theoretical yield" = 1.70 color(red)(cancel(color(black)("mol HNO"_3))) × ("1 mol NO")/(4 color(red)(cancel(color(black)("mol HNO"_3)))) = "0.425 mol HNO"_3"#

**4. Calculate the theoretical yield of #"NO"#**

#"Theoretical yield" = 0.425 color(red)(cancel(color(black)("mol NO"))) × ("30.01 g NO")/(1 color(red)(cancel(color(black)("mol NO")))) = "12.8 g NO"#