The balanced chemical equation between iron and oxygen to produce iron (III) oxide is,
Mass of Fe = 227.8 g
Moles of Fe =
Mass of oxygen = 128 g
Moles of
Calculating the limiting reactant: The reactant that produces the least amount of product will be the limiting reactant.
Mass of iron (III) oxide produced from Iron =
Mass of iron (III) oxide produced from oxygen=
Iron (Fe) produces the least amount of the product iron (III) oxide. So, Fe is the limiting reactant.
Answer:
3.38%
Explanation:
Given that;
the mass of the whole milk sample = 100 g
volume of HCl = 100 mL = 0.1 L
molarity of HCl = 0.5 M
volume of NaOH = 34.50 mL = 0.0345 L
molarity of NaOH = 0.3512 M
Since we knew the molarity and volume of both HCl and NaOH; we can calculate their corresponding number of moles present.
So, number of moles of HCl = molarity of HCl × volume of HCl
number of moles of HCl = 0.5 M × 0.100 mL
= 0.05 mole
number of moles of NaOH = Molarity of NaOH × Volume of NaOH
number of moles of NaOH = 0.3512 M × 0.0345 L
= 0.012 mole
From the question, we can deduce that the number of HCl that is consumed by NH₃ is equal to the number of moles of HCl that is consumed by NaOH.
SO, number of moles of HCl consumed by NH₃ = Total moles of HCl - moles of HCl consumed by NaOH
= 0.05 mole - 0.012 mole
= 0.038 mole
However, to determine the mole of NH₃ present , we have:
number of moles of NH₃ present = number of moles of HCl consumed by NH₃ = 0.038
∴ the mass of Nitrogen with the molecular weight (14.0 g/mol) = 0.038 moles × 14.0 g/mol
= 0.530 g
Now, the percentage of Nitrogen can be calculated as;
%
the percentage of protein in the sample = CP × %age of N
where CP is given as 6.38
∴ the percentage of protein in the sample = 6.38 × 0.530%
the percentage of protein in the sample = 3.3814%
the percentage of protein in the sample = 3.38%