Answer:
3 moles of CO are needed
Explanation:
Given data:
Number of moles of CO used = ?
Mass of Fe produced = 112 g
Solution:
Chemical equation:
Fe₂O₃ + 3CO → 2Fe + 3CO₂
Number of moles of Fe:
Number of moles = mass/ molar mass
Number of moles = 112 g/ 55.85 g/mol
Number of moles = 2.00 mol
Now we will compare the moles of iron and CO.
Fe : CO
2 : 3
Thus, 3 moles of CO are needed.
Answer:
Explanation:
Hello,
In this case, we write the reaction again:
In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:
Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:
But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:
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Answer:
The answer will be 2.98K
Explanation:
Using the formula:
Q = mc∆T
Q= 5,800 (heat in joules)
m= convert 15.2kg to g which is 15200g (mass in grams)
c= 0.128 J/g °c (Specific heat capacity)
∆T= what we need to find (temperature change)
5800J = 15200g x 0.128 x ∆T
= 2.98K
Answer:
F2 is the limiting reactant
27.6 grams of NaF is produced.
Explanation:
Balance the equation first.
2Na+ F2 ---> 2NaF
To find the limiting reactant, solve for how much NaF can be produced with Na and F2
12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)
=0.658 moles NaF
16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)
=0.705 moles NaF
Since F2 produced the least NaF, F2 is the limiting reactant.
Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.
0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF
27.6 moles of NaF would be theoretically produced.
Answer:
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