Question

Which reactant, and how many grams of it, is left over after 16.0 g of MnO2 (FW = 86.9 g/mol) and 30.0 g of HCl (FW = 36.5 g/mol) react according to the following chemical equation?MnO2 + 4 HCl ® MnCl2 + Cl2 + 2 H2O3.1 g HCl23.3 g HCl4.02 g MnO28.0 g MnO212.1 g MnO2

Answer 1

Answer : The HCl reactant is left over in the reaction and the left over amount of HCl is, 3.1 grams

Explanation : Given,

Mass of
`MnO_2` = 16.0 g

Mass of
`HCl` = 30.0 g

Molar mass of
`MnO_2` = 87 g/mole

Molar mass of
`HCl` = 36.5 g/mole

First we have to calculate the moles of
`MnO_2` and
`HCl`.

`\text{Moles of }MnO_2=\frac{\text{Mass of }MnO_2}{\text{Molar mass of }MnO_2}=(16g)/(87g/mole)=0.184moles`

`\text{Moles of }HCl=\frac{\text{Mass of }HCl}{\text{Molar mass of }HCl}=(30g)/(36.5g/mole)=0.822moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`MnO_2+4HCl\rightarrow MnCl_2+Cl_2`

From the balanced reaction we conclude that

As, 1 moles of
`MnO_2` react with 4 mole of
`HCl`

So, 0.184 moles of
`MnO_2` react with
`4* 0.184=0.736` moles of
`HCl`

From this we conclude that,
`HCl` is an excess reagent because the given moles are greater than the required moles and
`MnO_2` is a limiting reagent and it limits the formation of product.

The excess moles of HCl = 0.822 - 0.736 = 0.086 mole

Now we have to calculate the mass of
`HCl`.

`\text{Mass of }HCl=\text{Moles of }HCl* \text{Molar mass of }HCl`

`\text{Mass of }HCl=(0.086mole)* (36.5g/mole)=3.139g\approx 3.1g`

Therefore, the HCl reactant is left over in the reaction and the left over amount of HCl is, 3.1 grams