At 273.15 K and 100kPa, 58.34 g of #"HCl"# reacts with 0.35 mol of #"MnO"_2# to produce #"7.056 dm"^3# of chlorine gas. Calculate the theoretical yield of chlorine. How do I solve this?

Examine the balanced chemical equation for this redox reaction first.

Your objective at this stage is to determine which of the two reactants functions as a limiting reagent.

To ensure that every mole of manganese dioxide participates in the reaction, you must have

You can say that there will be excess hydrochloric acid because you have more moles than you need.

Because manganese dioxide will be totally consumed in the reaction, it will function as a limiting reagent.

The conditions that you were given for temperature and pressure are interesting because they match the most recent definition of STP.

You are aware that the response resulted in

of chlorine gas at STP, indicating that the reaction's real yield was

This can be used to determine the reaction's percent yield.

Enter your values to obtain

Two sig figs are used to round the answers.

Calculate moles of Cl2 produced: $n = \frac{volume}{molar\space volume}$

Find the limiting reactant: $\frac{moles\space given}{coefficient\space in\space the\space balanced\space equation}$

Use the limiting reactant to find moles of Cl2: $moles\space of\space limiting\space reactant \times coefficient\space of\space Cl2$

Convert moles of Cl2 to grams using molar mass: $moles\space of\space Cl2 \times molar\space mass\space of\space Cl2$