The balanced neutralization reaction here is:
Ca(OH)2 + 2HBr --> 2H2O + CaBr2
Notice that two moles of Her are required to neutralize every one mole of Ca(OH)2. This means that for however many moles of Her reacted, HALF as many moles of Ca(OH)2 reacted as well.
Moles of HBr reacted = 0.75 M x 0.345 L = 0.259 mol
Moles of Ca(OH)2 reacted = 0.259 mol / 2 = 0.130 mol
Concentration of Ca(OH)2 = 0.130 mol / 0.250 L = 0.52 M
Answer:
38.4 mol LiOH
Explanation:
Step 1: Write the balanced neutralization equation
2 LiOH + CO₂ ⇒ Li₂CO₃ + H₂O
Step 2: Establish the appropriate molar ratio
According to the balanced equation, the molar ratio of LiOH to CO₂ is 2:1.
Step 3: Calculate how many moles of lithium hydroxide are required to react with 19.2 mol CO₂
19.2 mol CO₂ × 2 mol LiOH / 1 mol CO₂ = 38.4 mol LiOH
The equation for this question could be 
→
.
so for 6.75 moles of 
*3 moles of O_{2}/2 moles of
= 10.125
