Question #6837f

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Question #6837f

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Answer 1 · 2016-03-29T23:48:49

The limiting reagent is the one that was only 0.237 moles. The other one is the excess reagent.

Explanation:

The carbonate is provided by the $N a_{2} C O_{3}$ $Na_2CO_3$ and the magnesium from the $M {g (N O_{3})}_{2} . 6 H_{2} O$ $Mg(NO_3)_2 . 6H_2O$ . ONLY the Mg and $C O_{3}$ $CO_3$ in the final product matter. First, a balanced equation is always required.

$M {g (N O_{3})}_{2} . 6 H_{2} O$ $Mg(NO_3)_2 . 6H_2O$ + $N a_{2} C O_{3}$ $Na_2CO_3$ → $M g C O_{3} (s) + 2 N O_{3}^{1 -} + 2 N a^{1 +}$ $MgCO_3 (s) + 2NO_3^(1-) + 2Na^(1+)$ (water is in solution)

Then calculate the number of moles in 2.00g $M g C O_{3}$ $MgCO_3$ . 2.00/84.3 = 0.0237. This is then also the total number moles of Mg and $C O_{3}$ $CO_3$ used. Whichever reactant was present in larger quantity (no quantities were specified in the question) is the excess reagent. The other one is the limiting reagent.

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Question #6837f

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