If 70.0g nitric acid reacts with 100.0g barium hydroxide what mass of salt may be obtained? If a student collects 138 g of this salt, what is the % yield?

Question

Natalie Anton · Accepted Answer

Theoretical maximum salt mass possible is #145,1529 g#.
Actual % yield obtained is #95%#.

The balanced chemical equation for this reaction, which is a neutralization/titration of acid and base, is #2HNO_3+Ba(OH)_2->Ba(NO_3)_2+2H_2O# We now convert all the reactants to moles because a balanced chemical equation shows the mole ratio in which chemicals react. #n_(HNO_3)=m/(M_r)=(70g)/((1+14+48)g//mol)=1,111 mol#. #n_(Ba(OH)_2)=m/(M_r)=100/(137,3+32+2)=0,584 mol#. Therefore #Ba(OH)_2# is in excess and #HNO_3# is the limiting reagent and controls how much of each product is formed. According to the balanced equation, 2 moles of the acid (nitric acid) produces 1 mole of the salt (barium nitrate). Hence by ratio and proportion, #1,111 mol# acid will produce # 0,5555 mol# of the salt. This will correspond to a mass of salt of #m=nxxM_r=0,5555xx(137,3+28+96)=145,1529#. Hence the percentage yield is #138/145.1529xx100=95%#.

Jeremiah Adams · Answer

undefined The balanced chemical equation for the reaction between nitric acid (HNO3) and barium hydroxide (Ba(OH)2) is:

2HNO3 + Ba(OH)2 → Ba(NO3)2 + 2H2O

The molar mass of nitric acid (HNO3) is 63.01 g/mol, and the molar mass of barium hydroxide (Ba(OH)2) is 171.34 g/mol.

1. Calculate the moles of nitric acid (HNO3) and barium hydroxide (Ba(OH)2) using the given masses:
   Moles of HNO3 = 70.0 g / 63.01 g/mol
   Moles of Ba(OH)2 = 100.0 g / 171.34 g/mol

2. Determine the limiting reactant by comparing the moles of both reactants.

3. Calculate the theoretical yield of barium nitrate (Ba(NO3)2) using the limiting reactant.

4. Calculate the percent yield using the actual yield (138 g) and the theoretical yield.