How many mole of H2 will be produced if 0.500 grams of magnesium is reacted with 10.00mL of 2.0 M HCl?
I've concluded the answer is 0.02 moles of H2, but I'm doubting myself a lot because I don't understand the aspects of the balanced chemical equation and how that affects the outcome of the answer. Here is my work, please tell me where I've gone wrong in my work, and so forth. Thanks.
~~~~~~ 2HCl + Mg ---> H2 + MgCl2 .02 mol HCl + .02057 mol Mg ---> 0.03mol MgCl2 +0.02mol H2 +0.01057mol Mg Mg Conversion to moles 0.5g X 1mol Mg/24.305g = 0.02057 mol Mg HCl Conversion to moles 2.0M HCl = n/0.010L = 0.02moles HCl Stoichiometry ~ 2HCl/1Mg = .02moles HCl/X X = .01 moles Mg will be consumed Left Over Mg .02057 mole Mg - .01 mole Mg = .01057moles Mg after reaction
Moles of MgCl2 .01 moles Mg + .02 moles Cl = .03 moles MgCl2 <-- I think this is wrong and should become .01 moles of MgCl2 because of the 2 on Cl.
moles of H2 Since all 10mL of 2.0M HCl is used up, then .02 moles of H2 is left. <-- I feel this is wrong because of the 2 of Hydrogen, so it should be .01 moles, not .02
Please help me in my confusion