Question

thanks for the help

Answer 1

is this the equation?

2 Mg+O2 --> 2 MgO

in that case, since moles of Mg = moles of MgO, convert grams of Mg to moles of Mg, then moles of Mg to moles of MgO, then moles of MgO to grams of MgO. that gives the theoretical yield

Answer 2

for percent yield, it's just

(actual yield/theoretical yield) * 100
where the actual yield is what's obtained in the experiment, and the theoretical yield is what's calculated

keep in mind you have to subtract the crucible/lid mass to get the actual yield

Answer 3

for average percent yield I think you would just average out the three percent yields you got from step 3

Answer 4

going from grams to moles, divide by the molar mass of Mg
going from moles to moles, you just need to multiply by the mole ratio (in this case, moles of Mg = moles of MgO so you just need to multiply by 1, which doesn't actually change the numer)
going from moles to grams, multiply by the molar mass of MgO

Answer 5

is it 24.18264 for trial 1?

Answer 6

hm I don't think so, remember it just wants how much MgO you get, not including the crucible or whatever

you have (26.993 - 26.682)g of Mg metal to start with (metal + crucible weight minus the crucible)

divide by the molar mass of Mg to convert to moles of Mg
multiply by the mole ratio of MgO/Mg (in this case just 1/1)
then multiply by the molar mass of MgO to get grams MgO

Answer 7

yeah

Answer 8

yup so that's your theoretical yield for Trial 1

Answer 9

yeah that looks good to me

Answer 10

thank you!