SAT II Chemistry : Stoichiometry with Reactions

Study concepts, example questions & explanations for SAT II Chemistry

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Example Questions

Example Question #22 : Sat Subject Test In Chemistry

Octane, \(\displaystyle C_{8}H_{18}\), is an important of component of gasoline, which can be burned as fuel in cars. The following is an unbalanced equation for the combustion of octane.

If 114 g of octane are burned in the complete combustion reaction shown below, how many grams of water will be produced?

___\(\displaystyle C_{8}H_{18} +\) ___\(\displaystyle O_{2} \rightarrow\) ___ \(\displaystyle CO_{2}\) \(\displaystyle +\) ___ \(\displaystyle H_{2}O\)

Possible Answers:

\(\displaystyle 324\)

\(\displaystyle 81\)

\(\displaystyle 162\)

\(\displaystyle 18\)

\(\displaystyle 180\)

Correct answer:

\(\displaystyle 162\)

Explanation:

First, the equation should be balanced. This can be done with the coefficients 2, 25, 16, and 18.

Next, calculating the molar mass of octane shows that it has a molar mass of 114 g/mol. This means that 1 mole of octane is used in the reaction, and 9 times as many moles of water should be produced, meaning one of the products will be 9 mol of water. Water has a molar mass of 18 g/mol, so the total mass of water produced is \(\displaystyle 9\cdot18=162\textup g\).

Example Question #11 : Stoichiometry

Suppose that 4 moles of \(\displaystyle NaClO_3\) undergoes complete decomposition. At STP, how many liters of \(\displaystyle O_2\) would be produced as product? (At STP, one mole of a gas takes up 22.4 L of space.)

Possible Answers:

\(\displaystyle 89.6\text{ L}\)

\(\displaystyle 44.8\text{ L}\)

\(\displaystyle 67.2\text{ L}\)

\(\displaystyle 134.4\text{ L}\)

\(\displaystyle 22.4\text{ L}\)

Correct answer:

\(\displaystyle 134.4\text{ L}\)

Explanation:

Start by creating a skeleton equation for the decomposition reaction:

\(\displaystyle NaClO_3 \rightarrow NaCl + O_2\)

This can be balanced to give:

\(\displaystyle 2NaClO_3 \rightarrow 2NaCl + 3O_2\)

This means that 4 mol of reactant would produce 6 mol of oxygen gas. The question states that at STP, one mole of a gas takes up 22.4 L of space, so it is easy to find that 6 moles of oxygen gas would occupy a volume of 134.4 L:

\(\displaystyle 6\text{ mol O}\times\frac{22.4 L}{1 \text{ mol}}=134.4\text{ L}\)

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