The reaction of calcium hydride and water produces calcium hydroxide and hydrogen as products. How many moles of \(\mathrm{H}_{2}(\mathrm{g})\) will be formed in the reaction between \(0.82 \mathrm{mol} \mathrm{CaH}_{2}(\mathrm{s})\) and \(1.54 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ?\)

Short Answer

Expert verified
1.54 moles of hydrogen gas will be produced.

Step by step solution

01

Analyze the stoichiometry of the reaction

It is clear from the balanced equation that for every mole of \(\mathrm{CaH}_{2}\), two moles of \(\mathrm{H}_{2}\) are generated.
02

Find the number of moles of hydrogen produced

Using stoichiometry, since we start with 0.82 moles of \(\mathrm{CaH}_{2}\), we will produce 0.82 moles × 2 = 1.64 moles of \(\mathrm{H}_{2}\) gas.
03

Compare with available moles of water

Ensure there is enough \(\mathrm{H}_{2}\mathrm{O}\) to react with the \(\mathrm{CaH}_{2}\). This is not limiting as we have 1.54 moles of \(\mathrm{H}_{2}\mathrm{O}\), and for every mole of \(\mathrm{CaH}_{2}\), only 2 moles of \(\mathrm{H}_{2}\mathrm{O}\) are required. 1.54 is less than 2×0.82, so water limits the reaction.
04

Calculate the number of moles of hydrogen produced when water limits the reaction

From every 2 moles of water, 2 moles of \(\mathrm{H}_{2}\) gas are produced. Hence, the amount of \(\mathrm{H}_{2}\) gas that is produced when water limits the reaction is 1.54 moles.

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Most popular questions from this chapter

A commercial method of manufacturing hydrogen involves the reaction of iron and steam. $$ 3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \stackrel{\Delta}{\longrightarrow} \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+4 \mathrm{H}_{2}(\mathrm{g}) $$ (a) How many grams of \(\mathrm{H}_{2}\) can be produced from \(42.7 \mathrm{g}\) Fe and an excess of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (steam)? (b) How many grams of \(\mathrm{H}_{2} \mathrm{O}\) are consumed in the conversion of \(63.5 \mathrm{g}\) Fe to \(\mathrm{Fe}_{3} \mathrm{O}_{4} ?\) (c) If \(14.8 \mathrm{g} \mathrm{H}_{2}\) is produced, how many grams of \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) must also be produced?

Consider the reaction below: \(2 \mathrm{AgNO}_{3}(\mathrm{aq})+\mathrm{Na}_{2} \mathrm{S}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{Ag}_{2} \mathrm{S}(\mathrm{s})+2 \mathrm{NaNO}_{3}(\mathrm{aq}) $$ (a) How many grams of \(\mathrm{Na}_{2} \mathrm{S}(\mathrm{s})\) are required to react completely with \(27.8 \mathrm{mL}\) of \(0.163 \mathrm{M} \mathrm{AgNO}_{3} ?\) (b) How many grams of \(\mathrm{Ag}_{2} \mathrm{S}(\mathrm{s})\) are obtained from the reaction in part (a)?

In the reaction shown, \(100.0 \mathrm{g} \mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}\) yielded \(64.0 \mathrm{g}\) \(\mathrm{C}_{6} \mathrm{H}_{10} .\) (a) What is the theoretical yield of the reaction? (b) What is the percent yield? (c) What mass of \(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}\) would produce \(100.0 \mathrm{g} \mathrm{C}_{6} \mathrm{H}_{10}\) if the percent yield is that determined in part (b)? $$\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{10}+\mathrm{H}_{2} \mathrm{O}$$

A method for eliminating oxides of nitrogen (e.g., \(\mathrm{NO}_{2}\) ) from automobile exhaust gases is to pass the exhaust gases over solid cyanuric acid, \(\mathrm{C}_{3} \mathrm{N}_{3}(\mathrm{OH})_{3}\) When the hot exhaust gases come in contact with cyanuric acid, solid \(\mathrm{C}_{3} \mathrm{N}_{3}(\mathrm{OH})_{3}\) decomposes into isocyanic acid vapor, HNCO(g), which then reacts with \(\mathrm{NO}_{2}\) in the exhaust gases to give \(\mathrm{N}_{2}, \mathrm{CO}_{2^{\prime}}\) and \(\mathrm{H}_{2} \mathrm{O}\) How many grams of \(\mathrm{C}_{3} \mathrm{N}_{3}(\mathrm{OH})_{3}\) are needed per gram of \(\mathrm{NO}_{2}\) in this method? [Hint: To balance the equation for reaction between HNCO and \(\mathrm{NO}_{2}\), balance with respect to each kind of atom in this order: \(\mathrm{H}, \mathrm{C}, \mathrm{O}, \text { and } \mathrm{N} .]\)

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