What is the molarity of 1.5 liters of an aqueous solution that contains 52 grams of lithium fluoride, LiF, (gram-formula mass =2
1 answer:
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Answer:
Are produced 12,1 g of AlCl₃ and 5,33 g of Al₂O₃ are left over
Explanation:
For the reaction:
Al₂O₃ (s) + 6 HCl (aq) → 2AlCl₃ (aq) + 3H₂0(l)
10,0g of Al₂O₃ are:
10,0g ₓ = <em>0,0980 moles</em>
And 10,0g of HCl are:
10,0 gₓ = <em>0,274 moles</em>
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For a total reaction of 0,274 moles of HCl you need:
0,274× = <em>0,0457 moles of Al₂O₃</em>
Thus, limiting reactant is HCl
The grams produced of AlCl₃ are:
0,274 moles HCl × × 133
= <em>12,1 g of AlCl₃</em>
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The moles of Al₂O₃ that don't react are:
0,0980 moles - 0,0457 moles =<em> </em>0,0523 moles
And its mass is:
0,0523 molesₓ = <em>5,33 g of Al₂O₃ </em>
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I hope it helps!