Question

Consider the reaction of zn metal with hydrochloric acid: zn(s) + 2hcl(aq) → zncl2(aq) + h2(g) if 2.57 g of zn is reacted with 0.500 moles of hcl in a 3.00 l container what pressure does the generated h2 exert against the container walls at 35.8 ℃?

Answer 1

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Answer 2

Answer: The pressure exerted by hydrogen gas is 0.33 atm

Explanation:

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

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

Given mass of zinc = 2.57 g

Molar mass of zinc = 65.4 g/mol

Putting values in above equation, we get:

$\text{Moles of zinc}=\frac{2.57g}{65.4g/mol}=0.039mol$

We are given:

Moles of HCl = 0.500 moles

For the given chemical equation:

$Zn+2HCl\rightarrow ZnCl_2+H_2$

By Stoichiometry of the reaction:

1 mole of zinc reacts with 2 moles of HCl

So, 0.039 moles of zinc will react with = $\frac{2}{1}\times 0.039=0.078mol$ of HCl

As, given amount of HCl is more than the required amount. So, it is considered as an excess reagent.

Thus, zinc metal is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of zinc reacts with 1 mole of Hydrogen gas

So, 0.039 moles of zinc will react with = $\frac{1}{1}\times 0.039=0.039mol$ of hydrogen gas

To calculate the pressure of hydrogen gas, we use the equation given by ideal gas:

PV = nRT

where,

P = Pressure of hydrogen gas = ?

V = Volume = 3.00 L

n = number of moles of hydrogen gas = 0.039 moles

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature = $35.8^oC=[35.8+273]K=308.8K$

Putting values in above equation, we get:

$P\times 3.00L=0.039mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 308.8K\\\\P=\frac{0.039\times 0.0821\times 308.8}{3.00}=0.33atm$

Hence, the pressure exerted by hydrogen gas is 0.33 atm