2Al + 6HCl --> 2AlCl3 + 3H2 Aluminium reacts with hydrochloric acid. How many grams of aluminum are necessary to produce 11 L
1 answer:
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Balanced equation: 2Na(s) + Cl₂(g) ---> 2NaCl(s)
when we have STP conditions, we can use this conversion: 1 mol = 22.4 L
first, we have to convert grams to molecules using the molar mass, and then use mole to mole ratio from the balanced equation.
molar mass of Na= 23.0 g/mol
ratio: 2 mol Na= 1 mol Cl₂ (based on coefficients of balanced equation)
calculations:
when we have STP conditions, we can use this conversion: 1 mol = 22.4 L
first, we have to convert grams to molecules using the molar mass, and then use mole to mole ratio from the balanced equation.
molar mass of Na= 23.0 g/mol
ratio: 2 mol Na= 1 mol Cl₂ (based on coefficients of balanced equation)
calculations:
Answer : The correct option is, (b) 22.1 g
Solution : Given,
Mass of iron = 15.5 g
Molar mass of iron = 56 g/mole
Molar mass of = 160 g/mole
First we have to calculate the moles of iron.
Now we have to calculate the moles of .
The balanced reaction is,
From the balanced reaction, we conclude that
As, 2 moles of iron obtained from 1 mole of
So, 0.276 moles of iron obtained from mole of
Now we have to calculate the mass of
Therefore, the amount of required are, 22.1 grams.