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22 March, 10:34

If you reacted 88.9 g of ammonia with excess oxygen, what mass of nitric oxide would you expect to make? You will need to balance the equation first. NH3 (g) + O2 (g) - > NO (g) + H2O (g)

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  1. 22 March, 11:02
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    1) Balanced chemical equation

    4NH3 (g) + 5O2 (g) - - - > 4NO (g) + 6H2O (g)

    2) State the molar ratios

    4 mol NH3 : 5 mol O2 : 4 mol NO : 6 mol H2O

    3) Convert 88.9 g of ammonia to moles, using the molar mass

    molar mass of NH3 = 14 g/mol + 3 * 1 g/mol = 17 g/mol

    number of moles = mass in grams / molar mass = 88.9 g / 17 g/mol = 5.23 mol NH3

    4) Make the proportion

    4 mol NO / 4 mol NH3 = x / 5.23 mol NH3=> x = 5.23 mol NO

    5) Convert 5.23 mol NO to grams

    molar mass NO = 14 g/mol + 16g/mol = 30 g/mol

    mass = number of moles * molar mass = 5.23 mol * 30 g/mol = 156.9 g ≈ 157 g

    Answer: 157 grams
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