Question #30c24

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Question #30c24

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Answer 1 · 2015-04-19T19:08:15

Your tool of choice for stoichiometry problems is the mole ratio.

For your balanced chemical equation, you have

$2 N H_{3 (g)} + 7 O_{2 (g)} \to 4 N O_{2 (g)} + 6 H_{2} O_{(l)}$ $color(red)(2)NH_(3(g)) + color(blue)(7)O_(2(g)) -> color(green)(4)NO_(2(g)) + 6H_2O_((l))$

Notice that you need $2$ $color(red)(2)$ moles of ammonia to react with $7$ $color(blue)(7)$ moles of oxygen gas in order to produce $4$ $color(green)(4)$ moles of nitrogen dioxide and 6 moles of water.

For the compounds involved in your reaction, you will always have the same ratios between the number of moles of each that react.

So, if you have 5.64 moles of oxygen that react, you'd need

$5.64 moles O_{2} \cdot \frac{2 moles N H_{3}}{7 moles O_{2}} = 1.61 moles$ $5.64cancel("moles "O_2) * (color(red)(2)" moles "NH_3)/(color(blue)(7)cancel("moles "O_2)) = "1.61 moles "$ $N H_{3}$ $NH_3$

Likewise, if 3.27 moles of oxygen react, you will produce

$3.27 moles O_{2} \cdot \frac{4 moles N O_{2}}{7 moles O_{2}} = 1.87 moles$ $3.27cancel("moles "O_2) * (color(green)(4)" moles "NO_2)/(color(blue)(7)cancel("moles "O_2)) = "1.87 moles "$ $N O_{2}$ $NO_2$

For the third scenario, you were given mass, not moles. To go from mass to moles, use ammonia's molar mass, which represents the mass of 1 mole of the substance

$8.95 g \cdot \frac{1 mole N H_{3}}{17.031 g} = 0.5255 moles$ $8.95cancel("g") * ("1 mole "NH_3)/(17.031cancel("g")) = "0.5255 moles "$ $N H_{3}$ $NH_3$

Once you have the moles, use the mole ratio that exists between ammonia and water

$0.5255 moles N H_{3} \cdot \frac{7 moles H_{2} O}{2 moles N H_{3}} = 1.84 moles$ $0.5255cancel("moles "NH_3) * ("7 moles "H_2O)/(color(red)(2)" moles "NH_3) = "1.84 moles "$ $H_{2} O$ $H_2O$

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Question #30c24

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