Multiple Choice
A sample of acetic acid is composed of 2.05 g of carbon and 2.73 g of oxygen. Calculate the mass of carbon in another sample of acetic acid that is composed of 3.27 g of oxygen.
3. Chemical Reactions
Stoichiometry
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According to the reaction Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H2S(g), what mass of Al2S3 remains unreacted when 20.00 g of Al2S3 and 2.00 g of H2O are reacted? The molar masses are: Al2S3 = 150.17 g/mol, H2O = 18.02 g/mol.
A
0.00 g
B
18.00 g
C
15.00 g
D
10.00 g
Verified step by step guidance1
Step 1: Begin by identifying the limiting reactant. To do this, calculate the number of moles of each reactant. Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). For Al2S3, \( \text{moles of Al2S3} = \frac{20.00 \text{ g}}{150.17 \text{ g/mol}} \). For H2O, \( \text{moles of H2O} = \frac{2.00 \text{ g}}{18.02 \text{ g/mol}} \).
Step 2: Use the stoichiometry of the balanced chemical equation to determine the mole ratio between Al2S3 and H2O. According to the equation, 1 mole of Al2S3 reacts with 6 moles of H2O. Calculate how many moles of H2O are required to completely react with the available moles of Al2S3.
Step 3: Compare the moles of H2O required with the moles of H2O available. If the available moles of H2O are less than the required moles, H2O is the limiting reactant. Otherwise, Al2S3 is the limiting reactant.
Step 4: If H2O is the limiting reactant, calculate how much Al2S3 reacts with the available H2O. Use the mole ratio from the balanced equation to find the moles of Al2S3 that react. Then, convert these moles back to grams using the molar mass of Al2S3.
Step 5: Subtract the mass of Al2S3 that reacts from the initial mass of Al2S3 to find the mass of Al2S3 that remains unreacted.
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