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50.0 kg of N_(2(g)) and 10.0 kg of H_(2(g)) are mixed to produce NH_(3(g)). Calculate the NH_(3(g)) formed. Identify the limiting reagent in the production of NH_(3) in this situation. |
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Answer» `N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g))`. Mole of `N_(2)=50.0 kg N_(2) xx (1000 g N_(2))/(1 kg N_(2)) xx (1 "mol" N_(2))/(28.09g N_(2))` `= 17.86 xx 10^(2)` mol Mole of `H_(2) = 10.00 kg H_(2) xx (1000 g H_(2))/(1 kg H_(2)) xx (1 "mol" H_(2))/(2.016 g H_(2))` `= 4.96xx10^(3)` mol According to the equation ,1 mol `N_(2(g))` required 3 mol `(H_(2))`. For the reaction. Hence, for `17.86xx10^(2)` mol of `N_(2(g))` , the MOLES of `H_(2(g))` required would be : `= 17.86 xx 10^(2)` mol `xx(3 "mol"H_(2(g)))/(1 "mol"N_(2(g)))` `= 5.36xx10^(3)` mol `H_(2(g))` But we have only `4.96xx10^(3)` mol `H_(2)`. Hence diphydrogen is the limiting reagent in this CASE. So, `NH_(3(g))` would be formed only from that amount of available dihydrogen , i.e `4.96xx10^(3)` mol `H_(2(g))`. Since 3 mol `H_(2(g)), 2 mol NH_(3(g))` `4.96xx10^(3) mol H_(2(g)) xx (2 mol NH_(3(g)))/(3 mol H_(2(g)))` `= 3.30xx10^(3) ` mol `NH_(3(g))` `3.30xx10^(3)` mol `NH_(3(g))` is obtained. If they are to be converted to GRAMS, it is done as follows: 1 mol `NH_(3(g))= 17.0 g NH_(3 (g))` `:.3.30xx10^(3) mol NH_(3(g))xx(17.0 g mol NH_(3(g)))/(1 mol NH_(3(g)))` `= 3.30xx10^(3) xx17 g NH_(3(g))` `= 56.1 xx 10^(3) g NH_(3)` `= 56.1 kg NH_(3(g))` |
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