We've got the balanced equation
and we can see that each 1 mole of "N{a_2}C{O_3}" reacts with 1 mole of "Ca{(H{C_2}{H_3}{O_2})_2}" , so the limited reagent in our situation is "N{a_2}C{O_3}" (because "5.51[{\\text{mol]}} < 6.01[{\\text{mol]}}") . Next, we can see that each 1 mole of "N{a_2}C{O_3}" give us 2 moles of "NaH{C_2}{H_3}{O_2}" , thus in this situation will be produced
"{\\nu _{NaH{C_2}{H_3}{O_2}}}{\\text{ = 2}} \\cdot {\\nu _{N{a_2}C{O_3}}} = 2 \\cdot 5.51[{\\text{mol]}} = 11.02[{\\text{mol]}}"moles of "NaH{C_2}{H_3}{O_2}" .
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