/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 Calcium hydride, CaH \(_{2},\) r... [FREE SOLUTION] | 91Ó°ÊÓ

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Calcium hydride, CaH \(_{2},\) reacts with water to form hydrogen gas: $$\mathrm{CaH}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{H}_{2}(g)$$ This reaction is sometimes used to inflate life rafts, weather balloons, and the like, when a simple, compact means of generating \(\mathrm{H}_{2}\) is desired. How many grams of \(\mathrm{CaH}_{2}\) are needed to generate 145 \(\mathrm{L}\) of \(\mathrm{H}_{2}\) gas if the pressure of \(\mathrm{H}_{2}\) is 825 torr at \(21^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
To generate 145 L of H$_2$ gas at 825 torr and 21°C, we need approximately \(m_{CaH_2} = \frac{1}{2} \times \frac{(825/760 \mathrm{atm})(145 \mathrm{L})}{(0.0821 \mathrm{L\ atm/mol\ K})(294.15 \mathrm{K})} \times 42.10 \mathrm{g/mol}\), which evaluates to approximately 34.54 grams of CaH$_2$.

Step by step solution

01

Calculate moles of hydrogen gas using the Ideal Gas Law

The Ideal Gas Law is given by: \[PV = nRT\] Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin. We are given P in torr, which we need to convert to atm by dividing by 760 torr/atm. First, let's convert the pressure and temperature to the required units: P = 825 torr = 825/760 atm T = 21°C = 21 + 273.15 K = 294.15 K Next, rearrange the Ideal Gas Law to solve for the number of moles (n): \[n = \frac{PV}{RT}\] Then, plug in the given values: \[n_{H_2} = \frac{(825/760 \mathrm{atm})(145 \mathrm{L})}{(0.0821 \mathrm{L\ atm/mol\ K})(294.15 \mathrm{K})}\]
02

Calculate moles of CaH2 required using the balanced equation

The balanced chemical equation is: \[CaH_2(s) + 2 H_2O(l) \rightarrow Ca(OH)_2(aq)+ 2 H_2(g)\] According to this equation, one mole of CaH2 generates two moles of hydrogen gas. We can use this ratio to find the moles of CaH2 required: \[n_{CaH_2} = \frac{1}{2} n_{H_2}\] Now, substitute the moles of hydrogen gas (n_{H_2}) from Step 1 into the equation: \[n_{CaH_2} = \frac{1}{2} \times \frac{(825/760 \mathrm{atm})(145 \mathrm{L})}{(0.0821 \mathrm{L\ atm/mol\ K})(294.15 \mathrm{K})}\]
03

Convert moles of CaH2 to grams using the molar mass

To find the mass of CaH2 needed, we multiply the moles of CaH2 by its molar mass: 40.08 g/mol (Ca) + 2(1.01 g/mol) (H) = 42.10 g/mol \[m_{CaH_2} = n_{CaH_2} \times M_{CaH_2} = \frac{1}{2} \times \frac{(825/760 \mathrm{atm})(145 \mathrm{L})}{(0.0821 \mathrm{L\ atm/mol\ K})(294.15 \mathrm{K})} \times 42.10 \mathrm{g/mol}\] After evaluating this expression, we can find the mass of CaH2 required to generate 145 L of H2 under the given conditions.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is an equation that represents the relationship between the pressure, volume, temperature, and the number of moles of a gas. It is expressed as PV = nRT, where P stands for pressure, V for volume, n for the number of moles of gas, R for the ideal gas constant, and T for the absolute temperature in Kelvin. Understanding this law is crucial for students tackling gas-related problems, as it allows you to connect and manipulate these variables to find the missing one.

To apply the Ideal Gas Law, make sure all units are consistent. Pressure is often given in units such as torr or millimeters of mercury (mmHg), but for the law to work, it should be in atmospheres (atm). The volume should be in liters (L), and the temperature in Kelvin (K). The constant R, which is 0.0821 L atm/mol K, remains the same unless your pressure and volume units differ. When faced with converting pressure from torr to atm, remember that 760 torr equals 1 atm. Thus, you can convert the given pressure by dividing it by 760.
Stoichiometry Calculations
Stoichiometry calculations are a methodical way of figuring out the quantitative relationships, or the ratio of amounts, in a chemical reaction. It relies heavily on the balanced chemical equation, which serves as a recipe, telling you how much reactant you need to make a certain amount of product, or vice versa.

For example, if a chemical equation shows that 2 moles of hydrogen combine with 1 mole of oxygen to produce 2 moles of water, stoichiometry allows us to calculate how many moles of oxygen are required to react with a specific number of moles of hydrogen. In solving problems, always start with the balanced equation—this controls the conversion between reactants and products. Using the molar ratios, you can convert moles of one substance to moles of another. In the given exercise, the 1 to 2 molar ratio of calcium hydride to hydrogen gas is key to finding the moles of calcium hydride required.
Molar Mass Conversion
Converting between moles and grams is something you'll need to do often in chemistry, and this is where the concept of molar mass becomes vital. The molar mass is the weight in grams of one mole of any substance. It tells you how many grams are in one mole of that substance, which is critical in stoichiometry calculations.

To find the molar mass, look at the periodic table and sum up the atomic weights of each element in the compound, accounting for the number of atoms of each. For instance, calcium hydride (CaHâ‚‚) consists of one calcium atom (40.08 g/mol) and two hydrogen atoms (each 1.01 g/mol). Therefore, its molar mass is the sum of these weights: 40.08 g/mol (calcium) + 2(1.01 g/mol) (hydrogen) = 42.10 g/mol. When you know the moles of a substance needed as determined by stoichiometric calculations, you multiply this number by the substance's molar mass to know the mass in grams.
Gas Generation Reaction
In a gas generation reaction, a gas is one of the products formed from the reactants. This type of reaction is commonly used when a particular gas needs to be produced onsite due to storage or transport difficulties. An example is the generation of hydrogen gas, which is used in everything from laboratory experiments to inflating life rafts or weather balloons, as illustrated by our original exercise involving calcium hydride reacting with water.

These reactions are practical for remote operations or emergency situations. Our example reaction is quite advantageous because it involves solid and liquid reactants that react to form a solid and a gas, making the inflation process simple. This hydrogen generation is predictable and controllable by the amount of calcium hydride and water used, thanks to stoichiometric calculations based on the balanced reaction equation. It's important to understand the safety considerations as well, since gases like hydrogen are flammable and must be handled with care.

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Most popular questions from this chapter

Calculate each of the following quantities for an ideal gas: (a) the volume of the gas, in liters, if 1.50 mol has a pressure of 1.25 atm at a temperature of \(-6^{\circ} \mathrm{C} ; \mathbf{b}\) ) the absolute temperature of the gas at which \(3.33 \times 10^{-3}\) mol occupies 478 \(\mathrm{mL}\) at 750 torr; (c) the pressure, in atmospheres, if 0.00245 \(\mathrm{mol}\) occupies 413 \(\mathrm{mL}\) at \(138^{\circ} \mathrm{C} ;(\mathbf{d})\) the quantity of gas, in moles, if 126.5 \(\mathrm{L}\) at \(54^{\circ} \mathrm{C}\) has a pressure of 11.25 \(\mathrm{kPa}\) .

Suppose you are given two \(1-\) flasks and told that one contains a gas of molar mass 30 , the other a gas of molar mass 60 , both at the same temperature. The pressure in flask \(A\) is \(x\) atm, and the mass of gas in the flask is 1.2 \(\mathrm{g}\) . The pressure in flask \(\mathrm{B}\) is 0.5\(x\) atm, and the mass of gas in that flask is 1.2 \(\mathrm{g}\) . Which flask contains the gas of molar mass \(30,\) and which contains the gas of molar mass 60\(?\)

(a) Calculate the density of sulfur hexafluoride gas at 707 torr and \(21^{\circ} \mathrm{C}\) . (b) Calculate the molar mass of a vapor that has a density of 7.135 \(\mathrm{g} / \mathrm{L}\) at \(12^{\circ} \mathrm{C}\) and 743 torr.

You have a sample of gas at \(-33^{\circ} \mathrm{C}\) . You wish to increase the rms speed by a factor of \(2 .\) To what temperature should the gas be heated?

The metabolic oxidation of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) in our bodies produces \(\mathrm{CO}_{2},\) which is expelled from our lungs as a gas: $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)$$ (a) Calculate the volume of dry \(\mathrm{CO}_{2}\) produced at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and 0.970 atm when 24.5 \(\mathrm{g}\) of glucose is consumed in this reaction. (b) Calculate the volume of oxygen you would need, at 1.00 \(\mathrm{atm}\) and \(298 \mathrm{K},\) to completely oxidize 50.0 \(\mathrm{g}\) of glucose.

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