Question

In the reaction mg(s)+2hcl(aq) h2(g)+mgcl2(aq), how many grams of hydrogen gas will be produced from 125.0 ml of a 6.0m hcl in an excess of mg

Answer 1

6.0m(mol/kg) of HCl
125mL H2O = 0.125kg
6mol/kg = n mol/0.125kg, n = 0.75mol
When 0.75mol of HCl reacts, 0.75/2=0.375mol of H2 is produced. H2 = 2g/mol
So, 0.375mol H2 = 0.75g

Answer 2

Answer: The amount of hydrogen gas produced will be 0.75 g.

Explanation:

To calculate the number of moles from molarity, we use the equation:

$\text{Molarity}=\frac{\text{Number of moles of solute}}{\text{Volume of solution}}$

We are given:

Molarity of solution = 6 M

Volume of solution = 125 mL = 0.125 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$6mol/L=\frac{\text{Moles of HCl}}{0.125L}\\\\\text{Moles of HCl}=0.75mol$

For the given chemical reaction:

$Mg(s)+2HCl(aq.)\rightarrow H_2(g)+MgCl_2(aq.)$

As, magnesium is present in excess, it is known as excess reagent and HCl is considered as limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of hydrochloric acid produces 1 mole of hydrogen gas.

So, 0.75 moles of hydrochloric acid will produce = $\frac{1}{2}\times 0.75=0.375moles$ of hydrogen gas.

To calculate the amount of substance, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of Hydrogen gas = 0.375 moles

Molar mass of hydrogen gas = 2 g/mol

Putting values in above equation, we get:

$0.375mol=\frac{\text{Mass of hydrogen}}{2g/mol}\\\\\text{mass of hydrogen gas}=0.75g$  

Hence, the amount of hydrogen gas produced will be 0.75 g.