a. 34 mL; b. 110 mL
a. A tablet containing 150 Mg(OH)₂
Mg(OH)₂ + 2HCl ⟶ MgCl₂ + 2H₂O
<em>Moles of Mg(OH)₂</em> = 150 mg Mg(OH)₂ × [1 mmol Mg(OH)₂/58.32 mg Mg(OH)₂
= 2.572 mmol Mg(OH)₂
<em>Moles of HCl</em> = 2.572 mmol Mg(OH)₂ × [2 mmol HCl/1 mmol Mg(OH)₂]
= 5.144 mmol HCl
Volume of HCl = 5.144 mmol HCl × (1 mmol HCl/0.15 mmol HCl) = 34 mL HCl
b. A tablet containing 850 mg CaCO₃
CaCO₃ + 2HCl ⟶ CaCl₂ + CO₂ + H₂O
<em>Moles of CaCO₃</em> = 850 mg CaCO₃ × [1 mmol CaCO₃/100.09 mg CaCO₃
= 8.492 mmol CaCO₃
<em>Moles of HCl</em> = 8.492 mmol CaCO₃ × [2 mmol HCl/1 mmol CaCO₃]
= 16.98 mmol HCl
Volume of HCl = 16.98 mmol HCl × (1 mL HCl/0.15 mmol HCl) = 110 mL HCl
The amount of Al required will be 15.77 grams
<h3>Stoichiometric problem</h3>
First, the equation of the reaction:

The mole ratio is 2:3.
Mole of 63.0 g of FeO = 63/71.84 = 0.8769 moles
Equivalent moles of Al = 0.8769 x 2/3 = 0.5846 moles
Mass of 0.5846 moles Al = 0.5846 x 26.98 = 15.77 grams
More on stoichiometric problems can be found here: brainly.com/question/14465605
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Answer:
1
Explanation:
The full question is shown in the image attached.
If we look at the structure closely, we will discover that there is only one C=C double bond in the molecule.
It is worthy of note that hydrogen (H2) only reduces the C=C double bond. it does not affect any other bond in the molecule.
Hence, only one molecule of H2 is required to reduce one C=C bond present.
Answer:
(a) 1.92 moles of Bi produced.
(b) 80.6 grams
Explanation:
Balanced equation: Bi2O3(s) + 3C(s) → 2Bi(s) + 3CO(g)
1st find moles of Bi2O3:
Bi2O3 has Mr of 466 and mass of 447 g


2nd find moles of Bi:
Bi2O3 : 2Bi
→ 1 : 2 ------ this is molar ratio.
→ 0.959227 : (0.959227)*2
→ 0.959227 : 1.91845
→ 0.959227 : 1.92
Therefore 1.92 moles of Bi was produced.
3rd Find moles of 3CO:
Bi2O3 : 3CO
1 : 3
0.959227 : (0.959227 )*3
0.959227 : 2.87768
3CO has 2.87768 moles and we know the Mr is 28.


g
Therefore 80.575 grams of CO was produced.