Stoichiometry: Quiz

Stoichiometry: Quiz

7 questions

1. Question 1EasyIn $\ce{2H2 + O2 -> 2H2O}$, what is the mole ratio of $\ce{O2}$ to $\ce{H2O}$?

   A1 : 1B1 : 2C2 : 1D16 : 18
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   Answer: 1 : 2

   The coefficients are 1 for $\ce{O2}$ and 2 for $\ce{H2O}$, so 1 mol of oxygen gives 2 mol of water. The answer 16 : 18 mixes up masses with moles.

2. Question 2EasyIn $\ce{N2 + 3H2 -> 2NH3}$, how many moles of $\ce{NH3}$ form from 3.0 mol of $\ce{H2}$ (with enough $\ce{N2}$)?

   A2.0 molB3.0 molC4.5 molD6.0 mol
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   Answer: 2.0 mol

   Mole ratio = want ÷ have = 2/3. So $3.0 \times \dfrac{2}{3} = 2.0$ mol. The answer 4.5 mol uses the ratio upside down (3/2).

3. Question 3EasyPropane burns as $\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$. How many moles of $\ce{O2}$ are needed to burn 2.00 mol of propane?

   A0.400 molB2.50 molC5.00 molD10.0 mol
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   Answer: 10.0 mol

   $2.00 \times \dfrac{5}{1} = 10.0$ mol of $\ce{O2}$. The answer 0.400 mol divides by 5 instead of multiplying.

4. Question 4MediumUsing the same equation, what mass of $\ce{CO2}$ forms when 1.00 mol of propane burns completely? ($M(\ce{CO2})$ = 44.01 g/mol)

   A44.0 gB88.0 gC132 gD220 g
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   Answer: 132 g

   1.00 mol of propane gives 3.00 mol of $\ce{CO2}$ (ratio 3/1). Mass = $3.00 \times 44.01 = 132$ g.

5. Question 5MediumWhat mass of ammonia forms from 28.0 g of nitrogen with excess hydrogen? $\ce{N2 + 3H2 -> 2NH3}$ (N = 14.01, H = 1.008)

   A17.0 gB28.0 gC34.0 gD56.0 g
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   Answer: 34.0 g

   Moles of $\ce{N2}$ = 28.0 ÷ 28.02 = 0.9993 mol. Moles of $\ce{NH3}$ = 0.9993 × 2 = 1.999 mol. Mass = 1.999 × 17.03 = 34.0 g. The answer 28.0 g wrongly assumes the masses are equal.

6. Question 6MediumIn $\ce{4Fe + 3O2 -> 2Fe2O3}$, how many moles of $\ce{Fe2O3}$ form from 6.00 mol of iron?

   A3.00 molB4.50 molC6.00 molD12.0 mol
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   Answer: 3.00 mol

   Mole ratio = want ÷ have = 2/4. So $6.00 \times \dfrac{2}{4} = 3.00$ mol of $\ce{Fe2O3}$.

7. Question 7HardA student uses the coefficients in $\ce{2H2 + O2 -> 2H2O}$ to say that 2 g of hydrogen reacts with 1 g of oxygen. What is wrong?

   ANothing; the coefficients give the mass ratioBThe coefficients give a mole ratio, not a mass ratioCThe equation is not balancedDHydrogen and oxygen do not react
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   Answer: The coefficients give a mole ratio, not a mass ratio

   Coefficients compare numbers of particles (moles). 2 mol of $\ce{H2}$ (4.03 g) reacts with 1 mol of $\ce{O2}$ (32.00 g), so the mass ratio is very different from 2 : 1.