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OpenStudy (ame2aw):
What volume of 0.3682 M H2SO4 solution is required to react with 0.4198 grams of Al(CN)3 according to the reaction, 2Al(CN)3+3H2SO4=6HCN+Al2(SO4)3?
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OpenStudy (anonymous):
You will need 15.20 ml approximately 210.08 gr of Al(CN)3 react with 294.21 |dw:1458877706245:dw| Then you only need to convert the mass into moles (0.58791/105.4 g) and divide the moles between the concentration to know how much volume you will need (Ans/0.3682 M) Correct me if I'm mistaken
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