Thermite's balanced equation threads through the whole SC180 Unit 6 stoichiometry problem set, from 10.0 g aluminum and 20.0 g iron(III) oxide to 14.0 g of iron. Searches like "sc 180 unit 6 assignment example", "sc180 unit 6 sample" and "sc180 unit 6 example" land here.
What a finished SC180 Unit 6 stoichiometry problem set looks like
Five linked problems on one reaction, 2Al plus Fe2O3 giving Al2O3 plus 2Fe. Problem one balances the equation and checks every element, not just iron. Problem two converts both starting masses to moles: 10.0 g over 26.98 g/mol gives 0.371 mol aluminum, and 20.0 g over 159.69 g/mol gives 0.125 mol oxide. Problem three tests the limiting reactant by asking how much aluminum the oxide would consume, 0.250 mol, less than the 0.371 available, so the oxide runs out first. Problem four carries the oxide's moles through the 1:2 ratio to 0.250 mol of iron and 14.0 g. Problem five reports an actual yield of 12.1 g as 86.5 percent and closes with a mass audit: 14.0 g iron, 12.8 g aluminum oxide and 3.24 g unreacted aluminum return the 30.0 g.
How a SC180 Unit 6 example is structured
The set is organized as a chain, so an early error would travel, and each problem restates the value it inherits before using it. Every calculation is written as a single line of factors, mass to moles, moles through the mole ratio, moles back to mass, with units canceling along the line. Molar masses are listed once in a small table at the top. The limiting reactant problem is written as an explicit comparison, required against available, with the conclusion stated in a sentence. Significant figures are decided from the given masses, three figures throughout, and the percent yield line shows actual over theoretical before multiplying by 100. The mass audit at the end is the set's own check: if reactant mass does not return as product plus leftover, some line above it holds the mistake.
Every element balanced, not only iron
The equation check lists aluminum, iron and oxygen with counts on each side, the habit that catches an equation balanced for the element the question names and no other.
Required against available
The oxide would consume 0.250 mol of aluminum and 0.371 mol is present, so aluminum is in excess. Writing the comparison out confirms the conclusion instead of assuming it from the smaller mass.
Grams to grams in a single line
The iron prediction is a single chain of three factors, grams of oxide canceling into moles and moles of iron canceling back into grams, so the ratio step stays visible.
Percent yield from the unrounded value
Actual yield of 12.1 g is divided by the unrounded theoretical 13.99 g, giving 86.5 percent. Dividing by the rounded 14.0 would give 86.4, and the set notes why that difference matters.
A mass audit that closes the set
Iron, aluminum oxide and leftover aluminum sum to the 30.0 g that went in, a conservation check that confirms the whole chain without repeating any calculation.
Where marks go in SC180 Unit 6
Sets lose most when the limiting reactant is chosen by comparing grams, which here names aluminum, the smaller mass, and gets it wrong. Mole ratios dropped or inverted come next, one mole of iron per mole of oxide in place of two, halving the prediction. Predicting iron from the excess aluminum gives 20.7 g, more iron than 20.0 g of oxide contains, and graders read that as a missed limiting step. An equation balanced for iron and left unbalanced for oxygen quietly corrupts every later line. A percent yield above 100, reported without comment, costs reasoning credit because it signals a calculation fault or an impure product. Moles given where mass was asked, or grams with no substance named, draw precision deductions.
Get a SC180 Unit 6 example written to your instructions
Copy out the Unit 6 problems for SC180, including the given masses, the reaction as printed and the rubric. Each problem is worked as a factor chain with the limiting reactant tested explicitly and yields reported to the right precision. The first worked set is free and returns within 24-48h.
SC180 Unit 6 questions, answered
Why can't I pick the limiting reactant by comparing masses?
Because reactions consume moles in a fixed ratio, not grams. Comparing masses here would name aluminum, the smaller amount, as limiting, and that is wrong: once both are converted to moles and the 2:1 ratio of aluminum to oxide is applied, the oxide runs out first. The comparison has to be required moles against available moles every time.
What does a percent yield over 100 percent mean?
In a paper problem it usually points to a ratio or molar mass error somewhere up the chain. In lab work, a product still wet or contaminated weighs more than it should. Either way, graders reward a sentence naming the likely cause over an answer that reports 108 percent without noticing that anything is wrong.
How many significant figures belong in a stoichiometry answer?
As many as the least precise measured quantity supports, which in this set is three from the given masses. Molar masses from the periodic table carry more digits and do not limit the result, and mole ratios from the balanced equation are exact. Carry extra digits through the chain and round only the final mass.