Chemistry help (1 Viewer)

[Farhanthestudent005]

Need help with this. Much appreciated if helped. Thanks

 

[icycledough]

For A), you need the equation first, which would be Mg + 2HCl --> MgCl2 + H2

Then, you need the moles of magnesium and the moles of hcl

n(Mg) = m(Mg) / mm(Mg) = 2.40 / 24.305 = 0.0987

n(HCl) = cV = 0.5 * 0.4 = 0.2

As question states it's difficult to see if magnesium is present, you can assume that it's been used up completely. If 0.0987 moles of magnesium is completely used up, there would be 1/2 the moles of HCl used up; i.e 0.2 / 2 = 0.1 ... as n(Mg) is less than n(HCl), you can infer that Magnesium is limiting reagent in this situation.

For B), using the reference sheet, at 25oC and 100kPA, the volume of 1 mole of an ideal gas is 24.79L. Thus, using the moles of magnesium to calculate the moles of gas produced (as Mg is limiting reagant), the equation tells us that the ratio of moles of Mg to H2 is 1:1 ... thus, n(H2) = 0.0987, and so the volume of gas would be 24.79L x 0.0987 = 2.44789138L = 2.48L (3s.f)

 

[CM_Tutor]

For A), you need the equation first, which would be Mg + 2HCl --> MgCl2 + H2

Then, you need the moles of magnesium and the moles of hcl

n(Mg) = m(Mg) / mm(Mg) = 2.40 / 24.305 = 0.0987

n(HCl) = cV = 0.5 * 0.4 = 0.2

As question states it's difficult to see if magnesium is present, you can assume that it's been used up completely. If 0.0987 moles of magnesium is completely used up, there would be 1/2 the moles of HCl used up; i.e 0.2 / 2 = 0.1 ... as n(Mg) is less than n(HCl), you can infer that Magnesium is limiting reagent in this situation.

Tweaking this slightly:

To use up all the Mg, the chemical amount of HCl required is 2 x 0.098745... = 0.19749... mol. The chemical amount of HCl present initially is 0.200 mol > 0.19749... mol. So, if all the Mg is used up, there is still HCl left and thus HCl is present in Xs / Mg is the limiting reagent.