Acompound of barium and oxygen was dissolved in hydrochloric acid to give a solution of barium ion, which was then precipitated with an excess of potassium chromate to give barium chromate. if a 1.345-g sample of the compound gave 2.012 g bacro4 , what is the formula of the compound?

The empiricial formula of the compound is BaO2

Explanation:

Step 1: Data given

Barium and oxygen dissolved in hydrochloric acid gives a solution of barium ion. This was precipitated with an excess of potassium chromate and gives barium chromate.

The original compound weighs 1.345g and gives 2.012g of BaCrO4

Step 2: Calculate moles of BaCrO4

moles of BaCrO4 = mass of BaCrO4 / molar mass of BaCrO4

moles of BaCrO4 = 2.012g / 253.37 g/mol = 0.0079 moles

Step 3: Calculate moles of Ba

Mole ratio for Ba and BaCrO4 is 1:1 so this means for 0.0079 moles of BaCrO4, there are 0.0079 moles of Ba-ion

Step 4: Calculate mass of Ba-ion

Mass of Ba = Moles of Ba / Molar mass of Ba

Mass of Ba = 0.0079 moles * 137.327g/mole = 1.085 g Ba

Step 5: Mass of oxygen

Since the original compound has barium and oxygen, the mass of oxygen is the difference between the original mass and the mass of the Ba-ion

1.345 g - (1.085 g Ba) = 0.260 g O

Step 6: Calculate moles of Oxygen

moles oxygen = mass of oxygen / Molar mass of oxygen

moles oxygen = 0.260g / 16g/mole = 0.01625 moles O

We divide the number of moles by the smallest number of moles which is 0.0079

Ba → 0.0079/0.0079 = 1

O → 0.01625 / 0.0079 ≈ 2

(1.091 g Ba) / (137.3277 g Ba/mol) = 0.00794450 mol Ba

(0.254 g O) / (15.99943 g O/mol) = 0.0158756 mol O

This gives us the empirical formula of BaO2 for this compound

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation: