Question

MnO2+ 4HCL--->MnCl2+Cl2+2H2O:
if 0.86 mole of MnO2 and 48.2g of HCL react, which reagent will be used up first? How many grams of Cl2 will be produced?
The answers are HCL, and 23.4g.
How would I get these answers?

Answer 1

since the equation is already balanced, you can proceed to solving for the limiting reagent

there are 0.86 moles of MnO2..so multiply this by the mole ratio. \[0.86 \; mol \; MnO_2 \times \frac{1 \; mole \; Cl_2}{1 \; mole \; MnO_2} \implies 0.86 \; \cancel{mol \; MnO_2} \times \frac{1 \; mole \; Cl_2}{1 \cancel{\; mole \; MnO_2}} = 0.86 \; mol \; Cl_2 \]
so for 0.86 moles of MnO2, there are 0.86 moles of Cl2

now solve HCl
\[48.2 \; g \; HCl \times \frac{1 \; mol \; HCl}{36.46 \; g \; HCl} \times \frac{1 \; mol \; Cl_2}{4 \; mol \; HCl} = 0.33 \; mol \; Cl_2\]
so since HCl produces lower number of mole of Cl2, HCl is the limiting reagent

Answer 2

now to get the amt of Cl2 produced, multiply 0.33 mol Cl2 by the molar mass of Cl2
\[\implies 0.33 \; mol \; Cl_2 \times \frac{70.9 \; g \; Cl_2}{1 \; mol \; Cl_2} = 23.397 \; g \; Cl_2\]

Answer 3

does that help @mistyl ?