Stoichiometry

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

Using the balanced equation, 2PbS + 3O₂ $$\rightarrow$$ 2PbO + 2SO₂, how many moles of PbO are made when .500 moles of O₂ are consumed?

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?

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?

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.

How many grams are in .333 moles of PbO? (Molar Mass: 223.2 g/mol)

If only 68.9 grams of lead (II) oxide are produced, but the reaction has a theoretical yield of 74.3 grams, what is the percent yield? $$\frac{Actual}{Theoretical}\times100\%$$

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

What is the mass of CO₂ produced?