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Problem 78

Aluminum hydroxide reacts with sulfuric acid as f…

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Problem 77

Sodium hydroxide reacts with carbon dioxide as follows:
$$
2 \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)
$$
Which is the limiting reactant when 1.85 mol $\mathrm{NaOH}$ and 1.00 $\mathrm{mol} \mathrm{CO}_{2}$ are allowed to react? How many moles of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ can be produced? How many moles of the excess reactant remain after the completion of the reaction?

Answer

See explanation for solution.


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Video Transcript

Okay, so we have a falling reaction off sodium hydroxide along with carbon dioxide to give us products. We know that one more off carbon dioxide reax hence carbon dioxide present in exist among any which is the limiting regent in this reaction for the now we need to calculate the moves off a Negussie or three produce that can be done as one for any two steel Tween derided by two moons off any for which multiplied with 1.85 moles off any well. Which giving US 0.9 to 5 mold off any too sealed three. Now the moles off carbon dioxide remained after completion. Off reaction can be determined. US. One move off, see or two minus 0.9 to 5 mourns off, See or two, which gives us 0.75 moles off. See you, too.

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