Question

Please help me with this question, and if u can do it on paper

Answer 1

1. O2 is the limiting reagent

2. Excess NH3 = 0.102 moles

3. 0.088 mol of NO is formed

4. 2.376 grams of H2O is formed

Stoichiometric problem

From the equation: $4NH_3 + 5O_2 --- > 4NO + 6H_2O$

Mole ratio of NH3 and O2 = 4:5

Mole of 3.25 g NH3 = 3.25/17 = 0.19 mol

Mole of 3.50 g O2 = 3.5/32 = 0.11 mol

Equivalent mole of O2 from NH3 = 5/4 x 0.19 = 0.24 mol

Equivalent mole of NH3 from O2 = 4/5 x 0.11 = 0.088 mol

Thus, O2 is the limiting reagent.

Excess NH3 = 0.19 - 0.088 = 0.102 moles

Mole ratio of O2 and NO = 5:4

Equivalent mole of NO = 4/5 x 0.11 = 0.088 mol

Mole ratio of O2 and H2O = 5:6

Equivalent mole of H2O = 6/5 x 0.11 = 0.132 mol

Mass of 0.132 mol H2O = 0.132 x 18 = 2.376 grams

More on stoichiometric problems can be found here: brainly.com/question/15047541

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