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Calcium hydride reacts with water to form calcium hydroxide and hydrogen gas. (a) Write a balanced chemical equation for the reaction. (b) How many grams of calcium hydride are needed to form \(4.500 \mathrm{~g}\) of hydrogen?

Short Answer

Expert verified
The balanced chemical equation for the reaction between calcium hydride and water is: \(CaH_2 + H_2O -> Ca(OH)_2 + 2H_2\) To form 4.500 g of hydrogen gas, we need 47.25 g of calcium hydride.

Step by step solution

01

Write the balanced chemical equation

The unbalanced chemical equation for the reaction between calcium hydride (CaH₂) and water (H₂O) is: CaH₂ + H₂O -> Ca(OH)₂ + H₂ Now we need to balance the equation. There are two hydrogen atoms in CaH₂ and two hydrogen atoms in H₂O, making a total of 4 hydrogen atoms. To balance the hydrogen atoms, we will put a coefficient of 2 in front of H₂: CaH₂ + H₂O -> Ca(OH)₂ + 2H₂ Now the equation is balanced with the same number of atoms for each element on both sides.
02

Calculate the moles of hydrogen gas

In order to find out how many grams of calcium hydride we need to form 4.500 g of hydrogen gas, we first need to calculate the moles of hydrogen gas. The molar mass of hydrogen gas (H₂) is approximately 2 g/mol. We can use the formula: moles = mass / molar_mass moles_of_H₂ = 4.500 g / 2 g/mol = 2.250 mol So, there are 2.250 moles of hydrogen gas.
03

Use stoichiometry to find the moles of calcium hydride

From the balanced chemical equation, we know that one mole of calcium hydride reacts to produce one mole of calcium hydroxide and two moles of hydrogen gas. This means that the molar ratio between CaH₂ and H₂ is 1:2. We can set up a proportion to find the moles of calcium hydride needed: moles_of_CaH₂ / moles_of_H₂ = 1 / 2 moles_of_CaH₂ = (1 × 2.250 mol) / 2 = 1.125 mol So, we need 1.125 moles of calcium hydride to produce 2.250 moles of hydrogen gas.
04

Calculate the mass of calcium hydride

Now that we know how many moles of calcium hydride we need, we can calculate the mass of calcium hydride required. The molar mass of CaH₂ is approximately 42 g/mol (40 g/mol for Ca and 2 g/mol for two hydrogen atoms). We can use the equation: mass = moles × molar_mass mass_of_CaH₂ = 1.125 mol × 42 g/mol = 47.25 g So, we need 47.25 grams of calcium hydride to produce 4.500 grams of hydrogen gas.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
Stoichiometry is a central concept in understanding chemical reactions because it allows us to make quantitative predictions about the outcomes of a chemical reaction.
It deals with the relationships between the amounts of reactants and products in a chemical reaction. By using stoichiometry, we can calculate masses, moles, and even concentrations of substances involved in reactions.
Here are some key points about stoichiometry:
  • It relies on the law of conservation of mass, meaning that matter is neither created nor destroyed in a chemical reaction.
  • Stoichiometric calculations are based on chemical equations, which must first be balanced.
  • The coefficients in a balanced chemical equation give the mole ratios needed to relate reactants to products.
Understanding stoichiometry involves mastering a few mathematical skills, including balancing equations and converting between masses and moles.
Balanced Chemical Equation
A balanced chemical equation is essential in stoichiometry as it shows the proportional relationship between reactants and products. Each element present in the reaction must have the same number on both sides of the equation, signifying no loss or gain of atoms. One crucial aspect of a balanced chemical equation is that it obeys the law of conservation of mass.
This is reflected in the fact that the total mass of reactants equals the total mass of products.
In our exercise, the equation for the reaction was:
  • Unbalanced: CaH₂ + H₂O → Ca(OH)₂ + H₂
  • Balanced: CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂
Balancing the chemical equation correctly helps to establish the stoichiometric coefficients needed for further calculations.
Molar Mass
Molar mass is a vital concept that connects the notion of mass with the amount of substance measured in moles.
It is defined as the mass of a given substance divided by the amount of substance. It is often expressed in grams per mole (g/mol).
For any element or compound, the molar mass can be calculated by adding up the atomic masses of all the atoms present. In our problem:
  • The molar mass of hydrogen gas (H₂) is approximately 2 g/mol.
  • The molar mass of calcium hydride (CaH₂) is approximately 42 g/mol (40 g/mol for calcium and 2 g/mol for two hydrogen atoms).
Understanding molar mass allows you to convert between grams and moles, which is essential for stoichiometric calculations in any chemical reaction.
Hydrogen Gas Production
In our exercise, hydrogen gas ( H₂) is one of the primary products, created during the reaction between calcium hydride and water.
Hydrogen gas is noted for being a clean-burning fuel and having applications in various industrial processes.
The production of hydrogen gas in chemical reactions can be predicted and calculated through stoichiometry:
  • By using the balanced equation, we know that two moles of hydrogen gas are produced for every one mole of calcium hydride consumed.
  • In practical applications, knowing the mass or volume of hydrogen produced allows for efficient reaction scaling and resource management.
Having knowledge about hydrogen gas production not only furthers practical applications, but also enhances our understanding of reaction dynamics and stoichiometry.

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Most popular questions from this chapter

Determine the formula weights of each of the following compounds: (a) Butyric acid, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH},\) which is responsible for the rotten smell of spoiled food; (b) sodium perborate, \(\mathrm{NaBO}_{3}\), a substance used as bleach; (c) calcium carbonate, \(\mathrm{CaCO}_{3},\) a substance found in marble. (c) \(\mathrm{CF}_{2} \mathrm{Cl}_{2},\) a refrigerant known as Freon; \((\mathbf{d}) \mathrm{NaHCO}_{3},\) known as baking soda and used in bread and pastry baking; \((\mathbf{e})\) iron pyrite, \(\mathrm{FeS}_{2}\) which has a golden appearance and is known as "Fool's Gold."

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