#1)a. Ba(NO3)2 + 2 NH2SO3H + 0 H2O = Ba(NH2SO3)2 + 2 HNO3 b. Ba(NO3)2 + NH2SO3H
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Question
#1)a. Ba(NO3)2 + 2 NH2SO3H + 0 H2O = Ba(NH2SO3)2 + 2 HNO3 b. Ba(NO3)2 + NH2SO3H + H2O = BaSO4 + NH4NO3 + HNO3 c. Ba(NO3)2 + 2 NH2SO3H + 2 H2O = Ba(NH2)2 + 2 H2SO4 + 2HNO3When a solution containing 1.4 g of Ba(NO3)2 and 2.4 g ofNH2SO3H is boiled, a precipitate forms. This precipitate maybe Ba(NH2SO3)2, BaSO4, or Ba(NH2) #2). Calculate the number of molesof each reactant. #3). Determine the limiting reactant in each of the following three reactions. The limiting reactant that you determine here will be used directly in this experiment. Reaction producing Ba(NH2SO3)2: Reaction producing BaSO4: Reaction producing Ba(NH2)2: #4). What conclusion can you reach about the limiting reactant in this experiment?
** I JUST NEED HELP ON 3 AND 4.
#1)a. Ba(NO3)2 + 2 NH2SO3H + 0 H2O = Ba(NH2SO3)2 + 2 HNO3 b. Ba(NO3)2 + NH2SO3H + H2O = BaSO4 + NH4NO3 + HNO3 c. Ba(NO3)2 + 2 NH2SO3H + 2 H2O = Ba(NH2)2 + 2 H2SO4 + 2HNO3
When a solution containing 1.4 g of Ba(NO3)2 and 2.4 g ofNH2SO3H is boiled, a precipitate forms. This precipitate maybe Ba(NH2SO3)2, BaSO4, or Ba(NH2) #2). Calculate the number of molesof each reactant. #3). Determine the limiting reactant in each of the following three reactions. The limiting reactant that you determine here will be used directly in this experiment. Reaction producing Ba(NH2SO3)2: Reaction producing BaSO4: Reaction producing Ba(NH2)2: #4). What conclusion can you reach about the limiting reactant in this experiment?
** I JUST NEED HELP ON 3 AND 4.
#1)a. Ba(NO3)2 + 2 NH2SO3H + 0 H2O = Ba(NH2SO3)2 + 2 HNO3 b. Ba(NO3)2 + NH2SO3H + H2O = BaSO4 + NH4NO3 + HNO3 c. Ba(NO3)2 + 2 NH2SO3H + 2 H2O = Ba(NH2)2 + 2 H2SO4 + 2HNO3
When a solution containing 1.4 g of Ba(NO3)2 and 2.4 g ofNH2SO3H is boiled, a precipitate forms. This precipitate maybe Ba(NH2SO3)2, BaSO4, or Ba(NH2) #2). Calculate the number of molesof each reactant. #3). Determine the limiting reactant in each of the following three reactions. The limiting reactant that you determine here will be used directly in this experiment. Reaction producing Ba(NH2SO3)2: Reaction producing BaSO4: Reaction producing Ba(NH2)2: #4). What conclusion can you reach about the limiting reactant in this experiment?
** I JUST NEED HELP ON 3 AND 4.
Explanation / Answer
Ba(NO3)2.
Figure out how many of each atoms you have in the compound and their atomic mass and multiply them and add them.
Ba ---> 1*137.3 =137.3
N ---> 2*14.0 = 28.0
O ---> 6*16.0 = 96.0
Total ---> 261.3 grams per mole.
But you only have 1.4 grams so just divide. 1.4/261.3 = 0.005 moles of Ba(NO3)2
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