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Chapter 3: Problem 66

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 of calcium hydride (CaH₂) with water (H₂O) is: CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂ To form 4.500 g of hydrogen gas, 46.889 g of calcium hydride are needed.

Step by step solution

01

Write the unbalanced chemical equation

Write the chemical equation for the reaction between calcium hydride (CaH₂) and water (H₂O), forming calcium hydroxide (Ca(OH)₂) and hydrogen gas (H₂): CaH₂ + H₂O → Ca(OH)₂ + H₂
02

Balance the chemical equation

Balance the chemical equation by placing coefficients in front of the chemical formulas to ensure equal numbers of atoms for each element on both sides of the equation: CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂
03

Calculate moles of hydrogen gas produced

To find out how many grams of calcium hydride are needed to form 4.500 g of hydrogen gas, first, we need to convert the mass of hydrogen gas into moles using the molar mass of hydrogen gas (2.02 g/mol): Moles of H₂ = \(\frac{4.500 \thinspace \text{g}}{2.02 \thinspace \text{g/mol}}\) = 2.228 moles H₂
04

Relate the moles of hydrogen gas to moles of calcium hydride

Using the balanced chemical equation, determine the stoichiometric relationship between moles of hydrogen gas and moles of calcium hydride: 1 mol CaH₂ → 2 mol H₂ Therefore, for every mole of calcium hydride, 2 moles of hydrogen gas are produced. Using this stoichiometric relationship, we can find the moles of calcium hydride required to produce 2.228 moles of hydrogen gas: Moles of CaH₂ = \(\frac{1 \thinspace \text{mol CaH}_{2}}{2 \thinspace \text{mol H}_{2}}\) × 2.228 moles H₂ = 1.114 moles CaH₂
05

Convert moles of calcium hydride to mass

Finally, convert the moles of calcium hydride into mass using the molar mass of calcium hydride (42.094 g/mol): Mass of CaH₂ = 1.114 moles CaH₂ × 42.094 \(\mathrm{\frac{g}{mol}}\) = 46.889 g CaH₂ So, 46.889 g of calcium hydride are needed to form 4.500 g of hydrogen gas.

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Most popular questions from this chapter

What parts of balanced chemical equations give information about the relative numbers of moles of reactants and products involved in a reaction?

Washing soda, a compound used to prepare hard water for washing laundry, is a hydrate, which means that a certain number of water molecules are included in the solid structure. Its formula can be written as \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot x \mathrm{H}_{2} \mathrm{O},\) where \(x\) is the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). When a \(2.558-\mathrm{g}\) sample of washing soda is heated at \(25^{\circ} \mathrm{C},\) all the water of hydration is lost, leaving \(0.948 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3} .\) What is the value of \(x ?\)

When ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) reacts with chlorine \(\left(\mathrm{Cl}_{2}\right)\), the main product is \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\), but other products containing \(\mathrm{Cl}\), such as \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2},\) are also obtained in small quantities. The formation of these other products reduces the yield of \(\mathrm{C}_{2} \mathrm{H}_{5}\) Cl. (a) Calculate the theoretical yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) when \(125 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{6}\) reacts with \(255 \mathrm{~g}\) of \(\mathrm{Cl}_{2}\), assuming that \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{Cl}_{2}\) react only to form \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}\) and HCl. (b) Calculate the percent yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) if the reaction produces \(206 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\).

What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{HCO}_{2},\) molar mass \(=90.0 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O},\) molar mass \(=88 \mathrm{~g} / \mathrm{mol}\).

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) together with other substances. Reaction of the ore with CO produces iron metal: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ (a) Balance this equation. (b) Calculate the number of grams of CO that can react with $$ 0.350 \mathrm{~kg} \text { of } \mathrm{Fe}_{2} \mathrm{O}_{3} $$ (c) Calculate the number of grams of Fe and the number of grams of \(\mathrm{CO}_{2}\) formed when \(0.350 \mathrm{~kg}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) reacts. (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.
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