Answer:
21.16 g
Step-by-step explanation:
Balance the equation
3 Zn + 2 MoO3 = Mo2O3 + 3 ZnO
And prepare their molar masses
Zn - 65.38
MoO3 - 143.96
Mo2O3 - 239.92
ZnO - 81.38
Since Zn is the limiting reagent; (you can determine this by trial and error but I'm too lazy), basically 29.2 g of MoO3 needs 19.892 g of Zn to react; while 17 g of Zn would need 24.955 g of MoO3 so we have shiet leftover. So we should always use the limiting reagent, in this case,
Zn
as it gets used up completely.
Anyway:
We can now solve using the ratio between Zn and ZnO.
(17 g of Zn /
65.38 g per Zn) x (3 mol of ZnO / 3 mol of Zn) x (81.38 g per ZnO / 1 mol of ZnO) = 21.16 g of ZnO
sooooo you get
21.16 g
(typing this on mobile and kinda hungover so yea)