Answer:
Explanation:
- For the balanced reaction:
<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s).</em>
It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.
- Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:
no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.
- Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:
<em><u>Using cross multiplication:</u></em>
4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.
0.64 mol of Fe is needed to react with → ??? mol of O₂.
∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.
- Finally, we can get the volume of oxygen using the information:
<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>
<em></em>
<em><u>Using cross multiplication:</u></em>
1 mol of O₂ occupies → 22.4 L, at STP conditions.
0.48 mol of O₂ occupies → ??? L.
∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.
Answer:
442.3 mL
Explanation:
Remember that Molarity is a measure of concentration in Chemistry and it's defined as the number of moles of the substance divided by liters of the solution:
Then, you can express 11.27 g of AgNO3 as moles of AgNO3 using the molar mass of the compound:
Then you can solve for the volume of the solution:
Hope it helps!
Answer:
Iodine is a mineral found in some foods. The body must have iodine to make thyroid hormones. The hormones control the body's metabolism and many other important functions. Without iodine, thyroid hormone production can decrease.
Explanation:
Answer: The answer can be found on CHEG
Explanation:
Answer:
4KNO3 ==> 2K2O + 2N2 + 5O2
Explanation:
It's a decomposition, but not a simple one.
KNO3 ==> K2O + N2 + O2 I don't usually do this, but I think the easiest way to proceed is to balancing the K and N together. That will require a 2 in front of KNO3
4KNO3 ==> 2K2O + 2N2 + 5O2
Now you have (3*4) = 12 oxygens. Two are on the K2O. So the other 10 must be on the O2
That should do it.
