To find out how many grams are in 4.65 moles of Al(NO₂)₃
Find out what the molar mass of Al(NO₂)₃ is
Al = 26.98 g/mol Al
N = 14 g/mol N
O = 16 g/mol O
Next, you have to look at the subscripts and figure out which they belong to, in this case:
Al = 26.98 g/mol Al
N₃ = 42 g/mol N₃
O₆ = 96 g/mol O₆
Finally, add the numbers together, so:
26.98 g/mol Al + 42 g/mol N₃ + 96 g/mol O₆ =
164.98 g/mol Al(NO₂)₃
Now, you have 4.65 mol Al(NO₂)₃ so
164.98 g/mol Al(NO₂)₃ × 4.65 mol Al(NO₂)₃ =
767.157 grams of Al(NO₂)₃
The wheels will be completely used up and it is the limiting reactant in this case.
<h3>What is a limiting reactant?</h3>
The limiting reactant is the reactant that is completely used up in a reaction, and thus determines when the reaction stops.
- 60 breaks will be used for 30 engines and 30 body frame
- 80 wheels will be used for 20 engines and 20 body frame
- 64 headlights will be used for 32 engines and 32 body frame
The wheels will be completely used up and it is the limiting reactant in this case.
Learn more about limiting reactants here: brainly.com/question/14222359
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<span>The force of a system can be measured by formula P=mf where P is the force, m is the mass of the system and f is the acceleration of the system. The formula is known as Newton's second law of motion.</span>
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