How many liters of oxygen are required to react completely with 14.8 mol of Al?
1 answer:
Answer:
296 L
Explanation:
We will need a balanced equation with moles, so let's gather all the information in one place.
4Al + 3O₂ ⟶ 2Al₂O₃
n/mol: 17.4
1. Moles of O₂
2. Volume of O₂
You haven't given the conditions at which the volume is measured, so I assume it is at STP (0 °C and 1 bar).
At STP the molar volume of a gas is 22.71 L.
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Answer:
1) 1.202 L
, 2) 1.291 dg
, 3) 204.877 and 4) 1.04x
Explanation:
You need to review about conversion factors and how to use them in the correct order. You can cancel the units and get the ones that you need if you use the appropriate conversion factors, remember is a number that you can use to multiply or divide.
For your exercise:
1) The conversion factor is: 1 L = 1000 mL
You will need to divide by 1000 mL to obtain liters L
1202.57120 mL x = 1.202 L
2) The conversion factor is: 1 g = 10 dg
0.1290743 g x = 1.291 dg
For the next exercises, you need to follow some rules:
1. All numbers that are different from Zero (non-zero digits) are significant figures.
2.The Zeros between non-zeros digits (Imbedded zeros) always are significant, 2007.
3. If you want to be specific and want some zeros to be significant you need to add a decimal point. For example 500. or 500.0
4. Leading zeros (to the left) are not significant.
5. Trailing zeros (zeros to the right) in a whole number without decimal point are not significant.
3) 843.062 - 638.1848 = 204.8772
Now if we round to 6 significant figures we get 204.877
4)123.0 x 8.43 = 1036.89
Now we round to 3 significant figures because 8.43 has the least significant figures.
1.04x