Ex 1:Hydrochloric acid reacts with magnesium hydroxide to producewater and aqueous magnesium chloride. How much water isproduced if 20.0 g of hydrochloric acid reacts with 10.0 g ofmagnesium hydroxide? How much of the excess reactant is leftover?
Question
Ex 1:Hydrochloric acid reacts with magnesium hydroxide to producewater and aqueous magnesium chloride. How much water isproduced if 20.0 g of hydrochloric acid reacts with 10.0 g ofmagnesium hydroxide? How much of the excess reactant is leftover?
Solution
To solve this problem, we need to follow these steps:
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Write the balanced chemical equation for the reaction. The equation is:
2HCl + Mg(OH)2 → 2H2O + MgCl2
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Calculate the number of moles for each reactant. The molar mass of HCl is approximately 36.5 g/mol and the molar mass of Mg(OH)2 is approximately 58.3 g/mol. So, we have:
Moles of HCl = 20.0 g / 36.5 g/mol = 0.547 moles Moles of Mg(OH)2 = 10.0 g / 58.3 g/mol = 0.172 moles
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Determine the limiting reactant. From the balanced equation, we can see that the reaction ratio of HCl to Mg(OH)2 is 2:1. So, we need twice as many moles of HCl for every mole of Mg(OH)2. But we have more than twice as many moles of HCl, so Mg(OH)2 is the limiting reactant.
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Calculate the amount of water produced. From the balanced equation, we can see that for every mole of Mg(OH)2, 2 moles of water are produced. So, we have:
Moles of H2O = 2 * 0.172 moles = 0.344 moles
The molar mass of water is approximately 18.0 g/mol, so the mass of water produced is:
Mass of H2O = 0.344 moles * 18.0 g/mol = 6.19 g
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Calculate the amount of excess reactant left over. We started with 0.547 moles of HCl and used up 0.172 * 2 = 0.344 moles in the reaction. So, we have:
Moles of HCl left = 0.547 moles - 0.344 moles = 0.203 moles
Converting this back to grams gives:
Mass of HCl left = 0.203 moles * 36.5 g/mol = 7.41 g
So, 6.19 g of water is produced and 7.41 g of HCl is left over.
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