The metal zinc will react with hydrochloric acid according to the following equation: Zn(s) + 2 HCl(aq) → H2(g) + ZnCl2(aq) If 100.0 g of zinc is mixed with 38.4 g of hydrochloric acid in water, how much hydrogen gas can be produced (in moles)?


Given data:

Zn(s)+2HCl(aq)⟶H2(g)+ZnCl2(aq)

100.0 g of zinc is mixed with 38.4 g of hydrochloric acid in water

Find: we need to calculate the number of moles of zinc and hydrochloric acid

Explanation:

we can use the stoichiometry of the balanced equation to find the moles of hydrogen gas produced.

Step 2 of 2

To determine the amount of hydrogen gas produced in moles, we need to calculate the number of moles of zinc and hydrochloric acid and then use the stoichiometry of the balanced equation to find the moles of hydrogen gas produced.

First, let’s calculate the number of moles of zinc (Zn):

Molar mass of Zn=65.38gmol−1

Number of moles of Zn=mass of Znmolar mass of Zn

Number of moles of Zn=100.065.38

Next, let’s calculate the number of moles of hydrochloric acid (HCl):

Molar mass of HCl=36.46gmol−1

Number of moles of HCl= mass oHClmolar mass of HCl

Number of moles of HCl=38.436.46

Now, let’s use the stoichiometry of the balanced equation to determine the moles of hydrogen gas produced.

According to the balanced equation:

1 mole of Zn reacts with 2 moles of HCl to produce 1 mole of H2

So, the moles of hydrogen gas produced will be equal to the moles of hydrochloric acid used.

Moles of H2=number of moles of HCl

Therefore, to find the moles of hydrogen gas produced, we can substitute the value for the number of moles of HCl into the equation:

Moles of H2=38.436.46

Calculating this expression:

Moles of H2=1.054mol

Explanation:

we have calculated the number of moles of zinc and hydrochloric acid and then use the stoichiometry of the balanced equation to find the moles of hydrogen gas produced.

Final solution

Therefore, 1.054moles of hydrogen gas can be produced from the reaction between 100.0g of zinc and 38.4g of hydrochloric acid.

 

 

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