Balanced chemical equation is:
2 Al + 6 HCl ---> 2 AlCl3 + 3 H2
Molar mass of Al = 26.98 g/mol
mass(Al)= 5.0 g
use:
number of mol of Al,
n = mass of Al/molar mass of Al
=(5 g)/(26.98 g/mol)
= 0.1853 mol
According to balanced equation
mol of HCl reacted = (6/2)* moles of Al
= (6/2)*0.1853
= 0.556 mol
This is number of moles of HCl
use:
M = number of mol / volume in L
0.3 = 0.556/ volume in L
volume = 1.85 L
volume = 1850 mL
Answer: 1850 mL
Question 11 of 20 Submit What volume (in mL) of 0.300 M HCl would be required...
What volume (in mL) of 0.300 M HCI would be required to completely react with 5.00 g of Al in the following chemical reaction? 2 Al(s) + 6 HCl(aq) → 2 AICI: (aq) + 3 H2 (g)
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