Could anyone help me with this chemistry equation??
Calculate the volume of chlorine needed to react with 14 g of iron .The equation for the reaction is 2Fe+3Cl2–->2FeCl3 Calculate the volume of chlorine needed to react with 14 g of iron Fe =56
Calculate the volume of chlorine needed to react with 14 g of iron .The equation for the reaction is 2Fe+3Cl2–->2FeCl3 Calculate the volume of chlorine needed to react with 14 g of iron Fe =56
Could anyone help me with this chemistry equation??
Calculate the volume of chlorine needed to react with 14 g of iron .The equation for the reaction is 2Fe+3Cl2–->2FeCl3 Calculate the volume of chlorine needed to react with 14 g of iron Fe =56
Do you know what the balanced equation means?
2Fe+3Cl2 ==> 2FeCl3
This says that 2 mol of iron will react with 3 mol of chlorine to produce 2 mol of iron(III) chloride. This ratio is ALWAYS respected.
Do you know how to calculate mol from mass? You need that skill for this question.
I understand the equation moles =mass/Rfm but I’m just not sure if I’ve got my number correct at all but I used 14 g /112 =0.125 mol and then I’m completely lost in what to do next even if that’s right which I don’t think it is ???
I understand the equation moles =mass/Rfm but I’m just not sure if I’ve got my number correct at all but I used 14 g /112 =0.125 mol and then I’m completely lost in what to do next even if that’s right which I don’t think it is ???
You should not use the equation coefficients when determining moles.
Moles of iron = 14/56 = 0.25 mol From the equation stoichiometry (the coefficients) you can see that 2 mol of iron makes 2 mol of iron(III) chloride.
So, however many mol of iron you have initially, you must end up with the same mol of iron(III) chloride. You start with 0.25 mol iron so you end up with 0.25 mol of iron(III) chloride.
The relative formula mass of iron(III) chloride = 56 x 3(35.5) = 162.5 So you end up with 0.25 x 162.5 g of iron(III) chloride = 40.625.
To know the mass of chlorine used you apply exactly the same process. Have a go!
You should not use the equation coefficients when determining moles.
Moles of iron = 14/56 = 0.25 mol From the equation stoichiometry (the coefficients) you can see that 2 mol of iron makes 2 mol of iron(III) chloride.
So, however many mol of iron you have initially, you must end up with the same mol of iron(III) chloride. You start with 0.25 mol iron so you end up with 0.25 mol of iron(III) chloride.
The relative formula mass of iron(III) chloride = 56 x 3(35.5) = 162.5 So you end up with 0.25 x 162.5 g of iron(III) chloride = 40.625.
To know the mass of chlorine used you apply exactly the same process. Have a go!
The answer I’ve got is 9dm3 does that sound correct