Lead will float in water.
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The limiting reactant is iron ore, the theoretical yield of iron metal is 701.344 kg, and the theoretical yield of carbon dioxide is 413.292 kg.
<h3>Stoichiometric problem</h3>From the equation of the reaction:
The mole ratio of iron ore to carbon is 2:3.
Mole of 1000 kg of iron ore = 1000000/159.69
= 6,262 moles
Mole of 120 kg carbon = 120000/12
= 10,000 moles
Thus, it appears that the carbon is in excess while the iron ore is limited in availability.
The mole ratio of the iron ore and the iron produced is 1:2. Thus, the equivalent number of moles of iron produced will be:
6,262 x 2 = 12,524 moles
Mass of 12,524 moles of iron = 12,524 x 56
= 701,344 g or 701.344 kg
Thus, the theoretical yield of iron is 701.344 kg.
The mole ratio of the iron ore and the carbon dioxide produced is 2:3. The equivalent mole of carbon dioxide produced will be:
6,262 x 3/2 = 9,393 moles
Mass of 9,393 moles carbon dioxide = 9,393 x 44
= 413,292 or 413.292 kg
The theoretical yield of carbon dioxide is, therefore, 413.292 kg.
More on stoichiometric problems can be found here: brainly.com/question/14465605
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Answer : The excess reactant is,
The leftover amount of excess reagent is, 7.2 grams.
Solution : Given,
Mass of = 105.0 g
Mass of = 78.0 g
Molar mass of = 80.11 g/mole
Molar mass of = 18 g/mole
Molar mass of = 100.09 g/mole
First we have to calculate the moles of and
.
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 1 mole of react with 3 mole of
So, 1.31 moles of react with
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles
Now we have to calculate the mass of excess reactant.
Thus, the leftover amount of excess reagent is, 7.2 grams.