Answer:
1040%
Explanation:
To solve this question we must convert the mass of Iron to moles in order to find limiting reactant. With limiting reactant we can find the theoretical moles of hydrogen and theoretical mass:
Percent yield = Actual yield (5.40g) / Theoretical yield * 100
<em>Moles Fe -Molar mass: 55.845g/mol-:</em>
10.3g * (1mol / 55.845g) = 0.184 moles of Fe will react.
For a complete reaction of these moles there are necessaries:
0.184 moles Fe* ( 3 mol H2SO4 / 2 mol Fe) = 0.277 moles H2SO4.
As there are 14.8 moles of the acid, <em>Fe is limiting reasctant.</em>
The moles of H2 produced are:
0.184 moles Fe* ( 3 mol H2 / 2 mol Fe) = 0.277 moles H2
The mass is:
0.277 moles H2 * (2.016g/mol) = 0.558g H2
Percent yield is:
5.40g / 0.558g * 100 = 1040%
It is possible the experiment wasn't performed correctly
First, find the number of moles of UF6
Avagadro's number = 6.023 x 10^23
Number of moles = 8.0 x 10^26 / Avagadro's number = 8.0 x 10^26 / 6.023 x 10^23 = 1.328 x 10³ moles
Molecular weight of UF6 = Molecular weight of U (238.02891) + Molecular weight of F6 (6 x 18.9984032) = 238.02891 + 113.9904192 = 352.0193292 g/mol
Therefore mass of 8.0 x 10^26 UF6 molecules = 352.0193292 g/mol x 1.328 x 10³ moles = 467.481669 x 10³ grams
Answer:
Reducing sugars are absent
Explanation:
Benedict's solution is an substance used in testing sugars. It is mixture of sodium carbonate, sodium citrate and copper(II) sulfate pentahydrate. It can be used instead of Fehling's solution in testing for the presence of reducing sugars.
Reducing sugars contain the -CHO group. If there is no colour change after the addition of Benedict's solution, then we can conclude that reducing sugars are absent.