Sulfuric acid dissolves aluminum metal according to the reaction: 2Al(s)+3H2SO4(aq)⇒Al2(SO4)3(aq)+3H2(g) Suppose you want to dissolve an aluminum block with a mass of 15.2 g. What minimum mass of H2SO4 (in g) do you need? What mass of H2 gas (in g) does the complete reaction of the aluminum block produce?

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Sulfuric acid dissolves aluminum metal according to the reaction:  2Al(s)+3H2SO4(aq)⇒Al2(SO4)3(aq)+3H2(g)  Suppose you want to dissolve an aluminum block with a mass of 15.2 g.  What minimum mass of H2SO4 (in g) do you need? What mass of H2 gas (in g) does the complete reaction of the aluminum block produce?

Virginia Palmer · Accepted Answer

Step 1  The balanced equation for the reaction between sulfuric acid (H2SO4) an aluminum (Al) is:  2Al(s)+3H2SO4(aq)⇒Al2(SO4)3(aq)+3H2(g)  Based on the balanced equation, 2 mol Al reacts with 3 mol H2SO4 to form 1 mol Al2(SO4)3 and 3 mol H2  Step 2  We can determine the amount of H2SO4 (in g) needed to dissolve 12.7 g Al by using the mole ratio their molar masses. Molar mass of Al=26.98gmol. Molar mass of H2SO4=98.08gmol  g H2SO4=12.7g Al×mol Al26.98g×3mol H2SO42mol Al×98.08gmol H2SO4  =69.3g H2SO4  Thus, 69.3 g H2SO4 is needed to dissolve a 12.7 g Al block  Step 3  We can determine the amount of H2 (in g) produced from the reaction between 69.3 g H2SO4 and 12.7 g Al using the mole ratio and the molar mass of H2 which is equal to 2.016gmol  g H2=12.7g Al×mol Al26.98g×3 mol H22 mol Al×2.016gmol H2  =1.43g H2  Therefore, 1.43 g H2 is produced from the rection between 69.3 g H2SO4 and 12.7 g Al.

godsrvnt0706 · Accepted Answer

Step 1  2Al(s)+3H2SO4(aq)⇒Al2(SO4)3(aq)+3H2(g)  Moles of Al=massmolar mass of Al  =15.626.98=0.5782mol  Moles of H2SO4=32×moles ofAl  =32×0.5782=0.8673mol  Mass of H2SO4=moles×molar mass ofH2SO4  =0.8673×98.08  =85.1g  Step 2  Moles of H2=32×moles ofAl  =32×0.5782=0.8673mol  Mass of H2=moles×molar mass ofH2  =0.8673×2.016  =1.75g

alenahelenash · Accepted Answer

Step 1 2Al(s)+3H2SO4(aq)⇒Al2(SO4)3(aq)+3H2(g) Moles of Al=massmolar mass of Al =15.226.98=0.5634mol Moles os H2SO4=32×moles ofAl =32×0.5634=0.8451mol Mass of H2SO4=moles×molar mass ofH2SO4 =0.8451×98.08=82.9g MSL Moles of H2=32×moles ofAl Mass os H2=moles×molar mass ofH2 =0.8451×2.016=1.70g

Sulfuric acid dissolves aluminum metal according to the reaction: 2Al(s)+3H_{2}SO_{4}(aq)\Rightarrow Al_{2}(SO_{4})_{3}(aq)+3H_{2}(g) Suppose

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