<li><p>Carminic acid, a naturally occurring red pigment extracted from the cochineal insect, contains only carbon, hydrogen, and oxygen. It was commonly used as a dye in the first half of the nineteenth century. It is 53.66% C and 4.09% H by mass. A titration required 18.02 mL of 0.0406M NaOH to neutralize 0.3602g carminic acid. Assuming that there is only one acidic hydrogen per molecule, what is the molecular formula of carminic acid?</p></li>
<li><p>When hydrochloric acid reacts with magnesium metal, hydrogen gas and aqueous magnesium chloride are produced. What volume of 5.0M HCl is required to react completely with 3.00g of magnesium?</p></li>
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<ol>
<li>What mass of solid aluminum hydroxide can be produced when 50.0mL of 0.200M Al(NO3)3 is added to 200.0mL of 0.100M KOH?</li>
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<p>Al(NO3)3 + KOH ---> Al(OH)3 + KNO3</p>
<p>50.0mL Al(NO3)3 x (1L/1000mL) x (1mol/22.4L) = .00223mol Al(NO3)3
200.0mL KOH x (1L/1000mL) x (1mol/22.4L) = .008929mol KOH</p>
<p>Therefore, the limiting reagent is Al(NO3)3. Then I found the mass of the product that can be produced with the limiting reagent.</p>
<p>.00223mol Al(NO3)3 x (1mol Al(OH)3/1mol Al(NO3)3) x (78.004g Al(OH)3/1mol Al(OH)3) = 1.74g Al(OH)3.</p>
<p>So it produces 1.74g Al(OH)3.</p>
<p>Is this solution right? Did I have to use the molarity given in the problem?</p>
<p>Can someone help me quick on this problem? I am stuck.</p>
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<li>A titration required 18.02mL of 0.0406M NaOH to neutralize 0.3602g carminic acid (empirical formula = C22H20O13). Assuming that there is only one acidic hydrogen per molecule, what is the molecular formula of carminic acid?</li>
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<p>Approach: I think I need to first find the molecular mass of carminic acid from the information given in the problem in order to find the molecular formula. I started to do that, but I don't know how to continue. Can someone please help me?</p>
<p>(0.0406mol NaOH/1L) x (1L/1000mL) x (18.02mL/1) x (39.998g NaOH/1mol NaOH) x .........</p>
<p>I don't know how to relate NaOH with carminic acid.</p>
<p>"49. What mass of solid aluminum hydroxide can be produced when 50.0mL of 0.200M Al(NO3)3 is added to 200.0mL of 0.100M KOH?"</p>
<p>The balanced equation should be</p>
<p>Al(NO3)3 + 3KOH = Al(OH)3 + 3KNO3</p>
<p>using molarity & volume, find the number of moles. use # of moles to find limiting reagent. There must be at least 3 times as much KOH as there is aluminum nitrate otherwise KOH is limiting reagent. Now, use the number of moles of limiting reagent to solve for grams of aluminum hydroxide. </p>
<p>22.4 L / mol is a characteristic of gases @ STP</p>