First, let's assign variables for the moles of CuSO₄·5H₂O as x and moles of MgSO₄·7H₂O as y. The molar mass of the substances are the following
CuSO₄·5H₂O: <span>249.685 g/mol
</span>CuSO₄: <span>159.609 g/mol
</span>MgSO₄·7H₂O: <span>246.49 g/mol
</span>MgSO₄: <span>120.366 g/mol
</span>H₂O: 18 g/mol
The solution is as follows:
2.988 = 249.685x + 246.49y --> eqn 1
5.02 - 2.988 = 18(5x + 7y) --> eqn 2
Solving both equations simultaneously, the values of x and y are:
x = 0.0134 mol CuSO₄·5H₂O
y = 0.0257 mol MgSO₄·7H₂O
Thus, the percent CuSO₄·5H₂O is equal to
Mass Percentage = [(0.0134 mol CuSO₄·5H₂O)*(249.685 g/mol )]/{[(0.0134 mol CuSO₄·5H₂O)*(249.685 g/mol )] + [(0.0257 mol MgSO₄·7H₂O)*(246.49 g/mol)]} * 100
Mass percentage = 34.56%
1) Divide the no. of atoms by Avogadros's no.
2) You will get no. Of moles . Now multiply the no. Of moles with the molar or molecular mass of Ag
3)The mass of Ag will be found ..
Answer:
B) Percent yield = 56%
Explanation:
Given data:
Mass of potassium iodide = 23 g
Mass of lead iodide formed = 18 g
Percent yield = ?
Solution:
Chemical equation:
2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂
Number of moles of potassium iodide:
Number of moles = mass / molar mass
Number of moles = 23 g/ 166 g/mol
Number of moles = 0.14 mol
Now we will compare the moles of PbI₂ and KI:
KI : PbI₂
2 : 1
0.14 : 1/2×0.14 = 0.07
Theoretical yield of PbI₂:
Mass = number of moles × molar mass
Mass = 0.07 × 461 g/mol
Mass = 32.27 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 18 g/ 32.27 g × 100
Percent yield = 56%