Mixed Practice Stoichiometry WS

Question

How many grams of silver bromide can be made from 34.7 g silver nitrate when silver nitrate and sodium bromide react? 

How many liters carbon dioxide gas are produced when 25.2 g of propane burns?

Hydrochloric acid reacts with magnesium to yield hydrogen gas and magnesium chloride. How many liters of hydrogen gas are produced when 0.18 moles of magnesium chloride are made?

How many grams of bromine can be made from 46.2 g sodium bromide when chlorine gas and sodium bromide react? 

____SO₂ + ____O₂  → ____SO₃

How many grams of oxygen are needed to produce 33.5 g of sulfur trioxide?

How many liters of oxygen gas will be needed to react with 8.19 x 10⁷ molecules of sulfur dioxide?

How many grams of sulfur dioxide will be needed to synthesize 1.01 moles of sulfur trioxide?

Answer 1

#1: AgNO₃ + NaBr --> AgBr + NaNO₃ [we can tell from this balanced reaction everything is a 1:1]

molar mass of AgNO₃ = (107.87g) + (14.01g) + (3*16g) = 169.88 g/mol

molar mass of AgBr = (107.87) + (79.9) = 187.77 g/mol

34.7g AgNO₃ x 1 mol AgNO₃ x 1 mol AgBr x 187.77g AgBr = 38.35 g AgBr

169.88g AgNO₃ 1 mol AgNO₃ 1 mol AgBr

# 2: C₃H₈ + 5O₂ --> 3CO₂ + 4H₂O

I am missing information such as the pressure and temperature (which need to be in units of atmospheres [atm] and Kelvin [K] respectively).

molar mass of C₃H₈ = (3*12.01g) + (8*1.01g) = 44.01 g/mol

25.2 g C₃H₈ x 1 mol C₃H₈ x 3 mol CO₂ = 1.72 mol CO₂

44.01 g C₃H₈ 1 mol C₃H₈

Now this is where we rearrange the ideal gas law to solve for L so --> L = nRT

P

#3: 2HCl + Mg --> H₂ + MgCl₂

Same idea for this problem, I can't solve fully due to lack of temperature and pressure.

molar mass of MgCl₂ = (24.31g) + (2*35.45) = 95.21 g/mol

0.18 mol MgCl₂ x 1 mol H₂ = 0.18 mol H₂ --> which can be used to plug in for n

1 mol MgCl₂

#4: Cl₂ + 2NaBr --> 2NaCl + Br₂

molar mass of NaBr = (23g) + (79.99g) = 102.99 g/mol

molar mass of Br₂ = (2*79.99) = 159.98 g/mol

46.2g NaBr x 1 mol NaBr x 1 mol Br₂ x 159.98 g Br₂ = 35.88 g Br₂

102.99g NaBr 2 mol NaBr 1 mol Br₂

#5: Using the reaction (which I balanced): 2SO₂ + O₂ --> 2SO₃

Go from:

1.) g of SO₃ --> g O₂

molar mass of SO₃ = (32.07g) + (3*16g) = 80.07 g/mol

molar mass of O₂ = (2*16g) = 32 g/mol

33.5 g SO₃ x 1 mol SO₃ x 1 mol O₂ x 32 g O₂ = 6.64 g O₂

80.07 g SO₃ 2 mol SO₃ 1 mol O₂

2.) Molec of SO₂ --> mol of O₂

8.19 x 10⁷ molec SO₂ x . 1 mol SO₂ x 1 mol O₂ = 6.80 x 10^-17 mol of O₂

6.022 x10²³ molec SO₂ 2 mol SO₂

3.) mol of SO₃ --> g of SO₂

molar mass of SO₂ = (32.07g) + (2*16g) = 64.07 g/mol

1.01 mol SO₃ x 2 mol SO₂ x 64.07 g SO₂ = 64.7 g SO₂

2 mol SO₃ 1 mol SO₂

AND DONE! ◡̈