Answer:
34.3 g
Explanation:
Step 1: Write the balanced equation
2 CH₃CH₂OH ⇒ CH₃CH₂OCH₂CH₃ + H₂O
Step 2: Calculate the moles corresponding to 50.0 g of CH₃CH₂OH
The molar mass of CH₃CH₂OH is 46.07 g/mol.
50.0 g × 1 mol/46.07 g = 1.09 mol
Step 3: Calculate the theoretical moles of CH₃CH₂OCH₂CH₃ produced
The molar ratio of CH₃CH₂OH to CH₃CH₂OCH₂CH₃ is 2:1. The moles of CH₃CH₂OCH₂CH₃ theoretically produced are 1/2 × 1.09 mol = 0.545 mol.
Step 4: Calculate the real moles of CH₃CH₂OCH₂CH₃ produced
The percent yield of the reaction is 85%.
0.545 mol × 85% = 0.463 mol
Step 5: Calculate the mass corresponding to 0.463 moles of CH₃CH₂OCH₂CH₃
The molar mass of CH₃CH₂OCH₂CH₃ is 74.12 g/mol.
0.463 mol × 74.12 g/mol = 34.3 g
Percent error is the difference between the measured and known value, divided by the known value, multiplied by 100%.
So first, we take our measured value, .299 cm, minus our known value, .225 cm.
.299 cm - .225 cm=.004 cm
Next, we divide that by our known value
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Finally, multiply your answer by 100
.0177777778 x 100= 1.77777778 %
Round to three significant figures, and you're done.
=1.78 % error
The mass of hydrogen atoms that is measured at 54 u given the relationship is 89.64×10¯²⁴ g
<h3>Conversion scale </h3>
1 u = 1.66×10¯²⁴ g
<h3>How to determine the mass of hydrogen atoms </h3>
- Mass of Hydrogen (u) = 54 u
- Mass of Hydrogen (g) =?
1 u = 1.66×10¯²⁴ g
Therefore
54 u = 54 × 1.66×10¯²⁴ g
54 u = 89.64×10¯²⁴ g
Thus, the mass of the hydrogen atoms measured at 54 u is 89.64×10¯²⁴ g
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Magnesium is a stronger reducing agent than copper and is thus able to reduce copper(II) oxide.
Products of the reaction: Magnesium oxide and metallic copper.