Answer:
Percent yield: 78.2%
Explanation:
Based on the reaction:
4Al + 3O₂ → 2Al₂O₃
<em>4 moles of Al produce 2 moles of Al₂O₃</em>
<em />
To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:
(Actual yield (6.8g) / Theoretical yield) × 100
Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:
4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.
As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:
0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = <em>0.0852 moles of Al₂O₃</em>,
In grams (Molar mass Al₂O₃ = 101.96g/mol):
0.0852 moles of Al₂O₃ × (101.96g / mol) =
<h3>8.7g of Al₂O₃ can be produced (Theoretical yield)</h3>
Thus, Percent yield is:
(6.8g / 8.7g) × 100 =
<h3>
78.2% </h3>
Answer:
Percent yield of SiC is 77.0%.
Explanation:
Balanced reaction: 
Molar mass of SiC = 40.11 g/mol
Molar mass of 
= 60.08 g/mol
So, 100.0 kg of = 
moles of = 1664 moles of
According to balanced equation, 1 mol of produces 1 mol of SiC
Therefore, 1664 moles of produce 1664 moles of SiC
Mass of 1664 moles of SiC = 
= 66743g = 66.74 kg (4 sig. fig.)
Percent yield of SiC = [(actual yield of SiC)/(theoretical yield of SiC)]
%
= 
%
= 77.0%
Answer:
1) 1.15 mol
2) M=0.45
3) 22.5 mL
4) 6.25 mL
Explanation:
1)
550 mL= 0.55 L
M= mol solute/ L solution
mol solute= M * L solution
mol solute= (2.1 M * 0.55 L ) M=1.15 mol solute
2)
155 mL = 0.155 L
80 g -> 1 mol NH4NO3
5.61 g -> x
x= (5.61 g * 1 mol NH4NO3)/80 g x= 0.07 mol NH4NO3
M=(0.07 mol NH4NO3)/0.155 L M=0.45
3) M1V1=M2V2
V1= M2V2/M1
V1= (0.500 M * 0.225 L)/5.00 M V1=0.0225 L =22.5 mL
4) M1V1=M2V2
V1= M2V2/M1
V1= (0.25 M * 0.45 L)/ 18.0 M
V1=6.25 x 10^-3 L = 6.25 mL
