
Calculate the number of grams of magnesium chloride that could be obtained from 17.0 g of HCl when HCl is reacted with excess magnesium oxide.
Answer
567.9k+ views
Hint: Start this question by writing the balanced chemical reaction of magnesium oxide and hydrochloric acid. Use the concept of molar mass and calculate weight of the individual compounds.
Complete step by step answer:
Let us solve this question by writing a balanced equation first.
According to the question, magnesium chloride is produced when hydrochloric acid reacts with excess of magnesium oxide. Therefore, the reaction can be written as –
\[MgO+2HCl\to MgC{{l}_{2}}+{{H}_{2}}O\]
Since excess of magnesium oxide is present, we can say that all 17.0 g of hydrochloric acid will be used up completely.
Molecular weight of HCl = 36.5 g
Since 2 moles of HCl is used, weight = 2 x 36.5 g = 73 g
Molecular weight of \[MgC{{l}_{2}}\] = molar mass of Mg + 2 x molar mass of Cl
Molecular weight of \[MgC{{l}_{2}}\] = 24 + 2 x 35.5 g = 95 g
Therefore,
73 g of HCl produces 95 g of \[MgC{{l}_{2}}\].
1 g of HCl produces \[\dfrac{95}{73}\]g of \[MgC{{l}_{2}}\].
17 g of HCl produces \[\dfrac{95}{73}\text{x17}\]g of \[MgC{{l}_{2}}\].
Weight of \[MgC{{l}_{2}}\] produced = \[\dfrac{95}{73}\text{x17}\]g = 22.12 g.
Therefore, the answer is – 22.19 g of magnesium chloride can be obtained from 17.0 g of HCl when HCl is reacted with excess magnesium oxide.
Note: The reaction - \[MgO+2HCl\to MgC{{l}_{2}}+{{H}_{2}}O\] is a neutralization reaction. In this, an acid reacts with an alkali to produce salt and water. Hydrochloric acid is an acid and magnesium oxide is a base, which reacts to produce a salt – magnesium chloride.
Complete step by step answer:
Let us solve this question by writing a balanced equation first.
According to the question, magnesium chloride is produced when hydrochloric acid reacts with excess of magnesium oxide. Therefore, the reaction can be written as –
\[MgO+2HCl\to MgC{{l}_{2}}+{{H}_{2}}O\]
Since excess of magnesium oxide is present, we can say that all 17.0 g of hydrochloric acid will be used up completely.
Molecular weight of HCl = 36.5 g
Since 2 moles of HCl is used, weight = 2 x 36.5 g = 73 g
Molecular weight of \[MgC{{l}_{2}}\] = molar mass of Mg + 2 x molar mass of Cl
Molecular weight of \[MgC{{l}_{2}}\] = 24 + 2 x 35.5 g = 95 g
Therefore,
73 g of HCl produces 95 g of \[MgC{{l}_{2}}\].
1 g of HCl produces \[\dfrac{95}{73}\]g of \[MgC{{l}_{2}}\].
17 g of HCl produces \[\dfrac{95}{73}\text{x17}\]g of \[MgC{{l}_{2}}\].
Weight of \[MgC{{l}_{2}}\] produced = \[\dfrac{95}{73}\text{x17}\]g = 22.12 g.
Therefore, the answer is – 22.19 g of magnesium chloride can be obtained from 17.0 g of HCl when HCl is reacted with excess magnesium oxide.
Note: The reaction - \[MgO+2HCl\to MgC{{l}_{2}}+{{H}_{2}}O\] is a neutralization reaction. In this, an acid reacts with an alkali to produce salt and water. Hydrochloric acid is an acid and magnesium oxide is a base, which reacts to produce a salt – magnesium chloride.
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