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Chlorine is prepared in the laboratory by treating manganese dioxide(MnO2) with aqueous hydrochloric acid according to the reaction
4 HCl + MnO2 -->2 H2O+ MnCl2 + Cl2
How many grams of HCl reacts with 5 grams of manganese dioxide ?​

Answer 1

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4 moles of HCl react with one mole of MnO₂ to give two moles of water and one mole of MnCl₂ and one mole of Cl₂ gas

4 moles of HCl weigh: 146g

1 mole of MnO₂ weighs: 87g

So 5 g MnO₂ will react with:   $\frac{87}{146} \times5g = \:\:8.39g\:\:of\:\:HCl$

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Answer 2

So the reaction is,

$\tt{\longrightarrow 4HCl+MnO_2\to2H_2O+MnCl_2+Cl_2}$

We get that 4 moles of $\tt{HCl}$ reacts with 1 mole of $\tt{MnO_2}$ to give the product.

$\tt{\longrightarrow 4\ mol\ HCl+1\ mol\ MnO_2\to Product}$

• Atomic mass of Hydrogen $\tt{=1\ g\,mol^{-1}.}$

• Atomic mass of Chlorine $\tt{=35.5\ g\,mol^{-1}.}$

• Atomic mass of Manganese $\tt{=55\ g\,mol^{-1}.}$

• Atomic mass of Oxygen $\tt{=16\ g\,mol^{-1}.}$

So in terms of mass,

$\tt{\longrightarrow 4(1+35.5)\ g\ HCl+1(55+2\times16)\ g\ MnO_2\to Product}$

$\tt{\longrightarrow 146\ g\ HCl+87\ g\ MnO_2\to Product}$

Multiplying the masses by $\tt{\dfrac{5}{87},}$

$\tt{\longrightarrow 146\times\dfrac{5}{87}\ g\ HCl+87\times\dfrac{5}{87}\ g\ MnO_2\to Product}$

$\tt{\longrightarrow 8.4\ g\ HCl+5\ g\ MnO_2\to Product}$

This means 8.4 grams of $\tt{HCl}$ reacts with 5 grams of $\tt{MnO_2}$ to give the product.