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SIZIF [17.4K]
2 years ago
8

What volume of 0.140 HCl is needed to neutralize 2.58 of Mg(OH)2 ? ...?

Chemistry
1 answer:
elena-s [515]2 years ago
8 0
To make a first step you have to know the balanced form for neutralization formula: Mg(OH)2(aq)+2HCl(aq)--\ \textgreater \ 2H2O+Mg(aq)+Cl(aq)

According to this, you can <span>calculate what you are being asked :</span>0.0442molMg(OH)2(x)(2molHCl)/(1molMg(OH)2)=0.0885molHCl
Then we have : 0.140MHCl=0.140(mol/L)HCl
Hope everything is clear, here is the exact answer you need : V=(0.0885molHCl)/(0.140(mol/L)HCl)=0.632LHCl&#10;

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4 0
2 years ago
How many moles of water are produced if 5.43 mol PbO2 are consumed?<br><br> ANSWER: 10.9
Arada [10]

Answer: 10.9 mol.

Explanation:

  • To understand how to solve this problem, we must mention the reaction equation where water produced from PbO₂.

Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O

  • Now, it is a stichiometric oriented problem, that 1 mole of PbO₂ produces 2 moles of H₂O.

Using cross multiplication:

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The moles of water produced = (5.43 x 2.0) = 10.86 moles ≅ 10.9 moles.

4 0
3 years ago
Read 2 more answers
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