Question

A student in this laboratory goes to the analytical balance and measures out 1.301 g of Ba(NO3)2 and 2.412 g of NH2SO3H. What mass in grams of Ba(NH2)2 can be produced according to the unbalanced chemical reaction below? If you need a periodic table of elements, please click HERE A A -... Ba(NO3)2 (aq) + .... NH2SO3(aq) + ---- H20(1) - _ -.. Ba(NH2)2 (s) + ---- H2504 (aq) + ---- HNO; (aq)

Answer 1

Balanced reaction :

15Ba(NO3)2 + 15NH2SO3H + 2H2O $\rightarrow$ 15Ba(NH2)2 + 20H2SO4 + 5HNO3

According to the reaction,

15 moles of NH2SO3H require 15 moles of Ba(NO3)2

1 mole NH2SO3H require 1 mole Ba(NO3)2

97 g NH2SO3H require 261 g Ba(NO3)2

2.412 g NH2SO3H require 261×2.412/97 g Ba(NO3)2

= 6.49 g Ba(NO3)2

Since we have less than this hence Ba(NO3)2 is limiting reagent.

According to the reaction,

15 mole Ba(NO3)2 give 15 mole of Ba(NH2)2

261 g Ba(NO3)2 give 169.37 g Ba(NH2)2

1.301 g Ba(NO3)2 give 169.37×1.301/261 g Ba(NH2)2

= 0.844 g Ba(NH2)2