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3 years ago

You have 100 grams of potassium fluoride (KF) .

Chemistry

1 answer:

laiz [17]3 years ago

6 0

Answer:

Option C = 1.72 mol

Explanation:

Given data:

Mass of KF = 100 g

Moles of KF = ?

Solution:

First of all we have to calculate the molar mass of KF.

Molar mass of KF = 39.0983 g/mol + 18.998403 g/mol

Molar mass of KF = 58. 0967 g/mol

Formula:

Number of moles = mass/molar mass

Number of moles = 100 g/ 58.0967 g/mol

Number of moles = 1.72 mol

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Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq)HCl(aq), as descr

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Answer: 0.887 g of $MnO_2$ should be added to excess HCl(aq).

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 805 torr = 1.06 atm (760torr=1atm)

V = Volume of gas = 235 ml = 0.235 L

n = number of moles = ?

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $25^0C=(25+273)K=298K$

$n=\frac{PV}{RT}$

$n=\frac{1.06atm\times 0.235L}{0.0820 L atm/K mol\times 298K}=0.0102moles$

$MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+2H_2O(l)+Cl_2(g)$

According to stoichiometry:

1 mole of chlorine is produced by = 1 mole of $MnO_2$

Thus 0.0102 moles of chlorine is produced by = $\frac{1}{1}\times 0.0102=0.0102$ moles of $MnO_2$

Mass of $MnO_2$ = $moles\times {\text {Molar mass}}=0.0102mol\times 87g/mol=0.887g$

0.887 g of $MnO_2$ should be added to excess HCl(aq).

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