Question

1. Given the balanced equation 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g), if 82.0 g of NH3 react with sufficient oxygen, how many grams of NO should form?

2. Consider the reaction between sodium metal and chlorine gas to form sodium chloride (table salt): 2Na(s) + Cl2(g) → 2NaCl(s) If the mass of the sodium solid increases by 0.500 g, what mass of sodium metal should have reacted?

Answer 1

1)

Molar mass of NH3,

MM = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

mass of NH3 = 82 g

mol of NH3 = (mass)/(molar mass)

= 82/17.03

= 4.814 mol

Balanced chemical equation is:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

According to balanced equation

mol of NO formed = moles of NH3

= 4.814 mol

Molar mass of NO,

MM = 1*MM(N) + 1*MM(O)

= 1*14.01 + 1*16.0

= 30.01 g/mol

mass of NO = number of mol * molar mass

= 4.814*30.01

= 144 g

Answer: 144 g

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