Balanced chemical equation for: Lead(II) sulfide, PbS, reacts with oxygen gas to produce lead(II) oxide, PbO, and sulfur dioxide, SO2.
Stoichiometry

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Chemistry
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10th Grade
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Hard
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1.
FLASHCARD QUESTION
Front
Back
2.
FLASHCARD QUESTION
Front
Back
2/3 as much
3.
FLASHCARD QUESTION
Front
Lead(II) sulfide, PbS, reacts with oxygen gas to produce lead(II) oxide, PbO, and sulfur dioxide, SO2. If 0.500 moles of O2 are used, what would the B row contain?
Back
.333 for PbS, 0 for PbO, and 0 for SO2
4.
FLASHCARD QUESTION
Front
Lead(II) sulfide, PbS, reacts with oxygen gas to produce lead(II) oxide, PbO, and sulfur dioxide, SO2. If 0.500 moles of O2 are used, what would the C row contain?
Back
-.333 for PbS, +.333 for PbO, and +.333 for SO2
5.
FLASHCARD QUESTION
Front
Lead(II) sulfide, PbS, reacts with oxygen gas to produce lead(II) oxide, PbO, and sulfur dioxide, SO2. If 0.500 moles of O2 are used, what would the A row contain? Options: 0 for PbS, .333 for PbO, and .333 for SO2; 0 for PbS, .750 for PbO, and .750 for SO2; .333 for PbS, .333 for PbO, and .333 for SO2; 0 for PbS, .500 for PbO, and .500 for SO2.
Back
0 for PbS, .333 for PbO, and .333 for SO2
6.
FLASHCARD QUESTION
Front
How many grams are in .333 moles of PbO? (Molar Mass: 223.2 g/mol)
Back
7.
FLASHCARD QUESTION
Front
Back
92.7%
8.
FLASHCARD QUESTION
Front
Which is the limiting reactant? Options: Oxygen gas will run out first, leaving .432 in excess CH4, Oxygen gas will run out first, leaving .863 in excess CH4, CH4 will run out first, leaving .432 in excess O2, CH4 will run out first, leaving .863 in excess O2
Back
Oxygen gas will run out first, leaving .432 in excess CH4
9.
FLASHCARD QUESTION
Front
What is the mass of CO2 produced?
Back
19.0 g of CO2
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