Lab 7: Limiting Reagent & Reaction Yields- Pre Lab Report 1. Answer the question

Question

Lab 7: Limiting Reagent & Reaction Yields- Pre Lab Report 1. Answer the questions concerning the chemical equation below 4 A (s)3 02 ( 2 Al-O, (s) a. Complete the mole relationships between each reactant and the product, with units (The mole relationship between the two reactants is 4 moes A to 3 mts) Al to Al:O O2 to Al:O b. Express the mole ratios from a as conversion factors, (ike the one on Page 2) There are four conversion factors, two for each ratio. How many grams of Al:0 will be produced from 163.0 grams of aluminum meta? (The formula weight of Al:0s is 102.0 g) Use the appropriate conversion factor from b for step two. Show the three steps of the calculation and be sure to include units and correct significant figures. c.

Explanation / Answer

a)

4 moles Al to 2 moles Al2O3

3 moles O2 to 2 moles Al2O3

b)

The mole ratio between Al and O2 is 4:3. For every 4 mole of Al used, 3 moles of O2 are consumed.

The mole ratio between Al and Al2O3 is 4:2 or 2:1

For every two moles of Al used, 1 mole of Al2O3 is formed.

C)

Step 1: moles present in 163.0 grams of Aluminium metal = mass of aluminium metal / atomic mass of aluminium

moles of aluminium = 163.0 g / 26.98 g/mol = 6.04 mol

Step 2: according to mole ratio between Al and Al2O3 from b) is 4:2 or 2:1

Step 3: moles Al2O3 formed = 6.04 / 2 mol = 3.02 moles

Step 4: convert moles Al2O3 into mass Al2O3 by multiplying with formula weight of Al2O3

Step 5: Grams of Al2O3 produced = 3.02 mol x 102.0 g/mol = 308 grams

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