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[40%" of a mixture of "0.2" mol of "N(2)...

[40%" of a mixture of "0.2" mol of "N_(2)" and "0.6mol" of "H_(2)" reacts to "],[" give "NH_(3)" according to the equation: "],[N_(2)(g)+3H_(2)(g)rightleftharpoons2NH_(3)(g)],[" at constant temperature and pressure.Then the ratio of "],[" he final volume to the initial volume of gases are "]

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