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Is the number of water molecules associated with each Al2(SO4)3
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The aluminum sulfate hydrate [Al2(SO4)3
. xH2O] contains8.10 percent Al by mass. Calculate x, that is, the number of water
Top answer:
decimal percent Al= (2*27)/(2*27+3*32+12*16 + x*18) so the decimal percent was given to be .0810,
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What would be the steps to solving this problem?
The aluminum sulfate hydrate [Al2(SO4)3 (times) xH2O] contains 8.20 % Al by
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RMM for Al2(SO4)3 =27*2+(32+4*16)*3 =342 RMM for H2O =2+16 =18 Number of water molecules = n Thus
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Hi, the question is: in each reaction, identify what have been oxidised and reduced.
The equation is: 2Al + 3H2SO4 ---> Al2(SO4)3
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Sulfuric acid exists in solution as H⁺ and HSO₄⁻ ions, but both react with Al metal to produce
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A sample of aluminum sulfate 18-hydrate, Al2(SO4)3 · 18 H2O, containing 157.0 mg is dissolved in 1.000 L of solution. Calculate
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I am looking for answer !!
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consider the fo9rmula for Al2(SO4)3,which is used in perspirants:
a. how many moles of sulfur are present i 3.0 moles of
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No need to count atoms/molecules. In 3.0 moles, there are 9.0 moles of S In .40 moles, there are .80
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Al2(SO4)3(aq) + BaCl2(aq) BaSO4(s) + AlCl3(aq)
Consider the unbalanced equation above. A 0.680 g sample of impure Al2(SO4)3
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To solve this problem, we need to determine the balanced equation for the reaction, and then use
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Al2(SO4)3(aq) + BaCl2(aq) BaSO4(s) + AlCl3(aq)
Consider the unbalanced equation above. A 0.680 g sample of impure Al2(SO4)3
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To find the mass percent of Al2(SO4)3 in the sample, we need to determine the amount of Al2(SO4)3
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Al2(SO4)3(aq) + BaCl2(aq) BaSO4(s) + AlCl3(aq)
Consider the unbalanced equation above. A 0.680 g sample of impure Al2(SO4)3
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To find the mass percent of Al2(SO4)3 in the sample, we need to determine the mass of Al2(SO4)3
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This is the equation for the formation of Al(OH)3:
6NaOH (aq) + Al2(SO4)3 (aq) �¨ 2Al(OH)3 (s) + 3Na2SO4 (aq) Would any side
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I don't think so. If too much NaOH is added, the Al(OH)3 dissolves with the formation of Al(OH)4^-
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consider the reaction, 2Al(OH)3+3H2SO4---->Al2(SO4)3+6H2O a. Given that 152g of H2SO4 reacts with 234g of Al(OH)3 Howmany grams
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I solve these limiting reagent problems by solving two simple stoichiometry problems, the first time
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