There are approximately 160 grams in 1 mol of fe203 molecules therefore there would be 79/160 = 0.49375 mols of fe203 molecules in 79 grams therefore 5 atoms in total for each molecule of fe203 therefore 79/160 *5 =79/32=2.46875 mols in atoms
I think its c, it makes sense to me to be c
2.22x10^-3 would be the answer to the question
please mark as brainliest answer
Answer: m = 50 g ZnSO4
Explanation: First is convert the moles of Zn to the moles of ZnSO4 by having their mole ratio which is 2:2 based from the balanced equation. Next is convert the moles of ZnSO4 to mass using its molar mass.
0.311 mole Zn x 2 moles ZnSO4 / 2 moles Zn
= 0.311 moles ZnSO4
0.311 moles ZnSO4 x 161 g ZnSO4 / 1 mole ZnSO4
= 50 ZnSO4