Answer:
The volume of CO2 produced is 6.0 L (option D)
Explanation:
Step 1: Data given
Volume of oxygen = 3.0 L
Carbon monoxide = CO = in excess
Step 2: The balanced equation
2 CO (g) + O2 (g) → 2 CO2 (g)
Step 3: Calculate moles of O2
1 mol of gas at STP = 22.4 L
3.0 L = 0.134 moles
Step 3: Calculate moles of CO2
For 2 moles CO we need 1 mol of O2 to produce 2 moles of CO2
For 0.134 moles O2 we'll have 2*0.134 = 0.268 moles CO2
Step 4: Calculate volume of CO2
1 mol = 22.4 L
0.268 mol = 22.4 * 0.268 = 6.0 L
The volume of CO2 produced is 6.0 L
Answer:
if i consider this reaction
Fe2O3+ 3CO---》2Fe+ 3CO2
so let's calculate first moles of Fe2O3 i.e. = 256/159.69= 1.6 moles
So the one moles of Fe2O3 is forming three moles of CO2
hence 1.6 moles will form 4.8 moles of CO2
one mole of CO2 is 44 g so 4.8 moles of Co2 is 44×4.8= 211.2 g
so the conclusion is 211.2 g of CO2 can be produced from 256 g Fe2O3!!
i d k it's right or wrong but i tried my best :)
I got this idk if it's correct: For measuring large distances, the kilometer (103 or 1000 meters) is often used. The basic unit of volume in the metric system is the liter (l). The most common derived unit is the milliliter (ml) (10-3 or 1/1000 of a liter). The volume of a milliliter is equal to the volume of a cube 1 centimeter per side.
Answer:
At the cathode during the electrolysis of an aqueous solution of magnesium iodide, MgI2 , 2I−(aq) is produced
Explanation:
At cathode, reduction reaction takes place.
The dissociation of MgI2 in aqueous solution is Mg2+(aq) and 2I−(aq)
Here, the Iodine reduces to 2I−(aq) from state of 0 (MgI2) to state of -1 (2I−(aq))
Hence, at the cathode during the electrolysis of an aqueous solution of magnesium iodide, MgI2 , 2I−(aq) is produced
