Answer:
Amount of elemental iron in sample A is 0.0405g.
Step-by-step explanation:
Molar mass of C = 12 g/mol
Molar mass of H = 1 g/mol
Molar mass of Fe = 56 g/mol
Molar mass of O = 16 g/mol
So, molar mass of ferrous gluconate =
![(12* 12)g+(24* 1)g+(1* 56)g+(14* 16)g=448g](https://img.qammunity.org/2020/formulas/chemistry/college/3qysteivaxqn0zwe59dl3qmu2xu5upi80g.png)
So number of mole of ferrous gluconate in 324 mg =
![(0.324)/(448)moles](https://img.qammunity.org/2020/formulas/chemistry/college/wgz9p3wenrrwopn2hl9hbg8vtzff2m2w8e.png)
(number of moles =
)
As 1 mol of ferrous gluconate contains 1 mol of Fe therefore
of ferrous gluconate contain
of Fe.
So amount of elemental iron in sample A =
![(0.324)/(448)* 56g=0.0405g](https://img.qammunity.org/2020/formulas/chemistry/college/lpxqm55frneomeh81p3x4p83wyjm9eeo8r.png)