2.)For the balanced equation: 2HNO 3 + Ba(OH) 2 Ba(NO 3 ) 2 + 2H 2 O A.) If you

ID: 571140 • Letter: 2

Question

2.)For the balanced equation: 2HNO3 + Ba(OH)2 Ba(NO3)2 + 2H2O

A.) If you had 55.00 g of HNO3 and enough Ba(OH)2, how many g of Ba(NO3)2 and moles of water would be produced?(1 point) Watch Sig Figs

B.) Based on your answer in 2a., if you actually produced 0.25 g of Ba(NO3)2, what is the % yield? (1 point)

C.) If you had 0.250 g of HNO3 and 0.350 g of Ba(OH)2, which one is the limiting agent. Show your work to support your answer. (1 point

*****Please explain Step by Step********

2HNO3 + Ba(OH)2 ------------------> Ba(NO3)2 + 2H2O

126 g 171.3 g 261.3g 36 g

55.0 g ?

mass of Ba(NO3)2 = 261.3 x 55 / 126

= 114.0g

mass of Ba(NO3)2 = 114.0g

mass of water = 55 x 36 / 126

= 15.7 g

moles of water = 15.7 / 18

= 0.873

moles of water = 0.873

% yield = actual x 100 / thereotical

= 0.25 x 100 / 114

= 0.219%

2HNO3 + Ba(OH)2 ------------------> Ba(NO3)2 + 2H2O

126 g 171.3 g 261.3g 36 g

0.250g 0.350 g

126 g HNO3 --------------------> 171.3 g Ba(OH)2

0.250 g HNO3 ------------------> 0.250 x 171.3 / 126 = 0.340 g Ba(OH)2 needed

but we have 0.350 g Ba(OH)2 so . it is in excess . Ba(OH)2 is the excess reagent

HNO3 is the limiting reagent

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.