Answer:
Mg(s)+2HCl(aq)→MgCL2(aq)+H2(g), what mass of hydrogen will be obtained if 100 cm3 of 2.00 mol dm−3 HCl are added to 4.86 g of magnesium?
Explanation:
Moles of metal, = 4.86⋅g24.305⋅g⋅mol−1 = 0.200 mol.
Moles of HCl = 100⋅cm−3×2.00⋅mol⋅dm−3 = 0.200 mol
Clearly, the acid is in deficiency ; i.e. it is the limiting reagent, because the equation above specifies that that 2 equiv of HCl are required for each equiv of metal.
So if 0.200 mol acid react, then (by the stoichiometry), 1/2 this quantity, i.e. 0.100 mol of dihydrogen will evolve.
So, 0.100 mol dihydrogen are evolved; this has a mass of 0.100⋅mol×2.00⋅g⋅mol−1 = ??g.
If 1 mol dihydrogen gas occupies 24.5 dm3 at room temperature and pressure, what will be the VOLUME of gas evolved?
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