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"#
