Answer:
1. 7.81 moles HgO
2. n = mass/molar mass = (4000 g)/(159.69 g/mol) = 25.05 mol.
Explanation:
How many moles of mercury (II) oxide are needed to produce 125 g of oxygen?
2HgO ==> 2Hg + O2
125 g O2 x 1 mol O2/32 g x 2 mol HgO / mol O2 = 7.81 moles HgO
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If 4000 g of Fe2O3 is available to react, how many moles of CO are needed?
The no. of moles of CO are needed = 75.15 mol.
Fe₂O₃ + 3CO → 2Fe + 3CO₂,
It is clear that 1 mol of Fe₂O₃ reacts with 3 mol of CO to produce 2 mol of Fe and 3 mol of CO₂.
If 4.00 kg Fe₂O₃ are available to react, how many moles of CO are needed?
We need to calculate the no. of moles of 4.00 kg Fe₂O₃:
n = mass/molar mass = (4000 g)/(159.69 g/mol) = 25.05 mol.
Using cross multiplication:
1 mol of Fe₂O₃ need → 3 mol of CO to react completely, from stichiometry.
25.05 mol of Fe₂O₃ need → ??? mol of CO to react completely.
The no. of moles of CO are needed = (3 mol)(25.05 mol)/(1 mol) = 75.15 mol.
