Question

If you start with 89.3 g no(g) and 28.6 g h2(g), find the theoretical yield of ammonia.

Answer 1

50.7 g. First, lookup the atomic weights of all the involved elements. Atomic weight nitrogen = 14.0067 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Now calculate the molar masses Molar mass NO = 14.0067 + 15.999 = 30.0057 g/mol Molar mass H2 = 2 * 1.00794 = 2.01588 g/mol Molar mass NH3 = 14.0067 + 3 * 1.00794 = 17.03052 g/mol Calculate how many moles of each reactant we have: Moles NO = 89.3 g / 30.0057 g/mol = 2.9761 mol Moles H2 = 28.6 g / 2.01588 g/mol = 14.1874 mol The balanced equation for the reaction is 2NO + 5H2 ==> 2NH3 + 2H2O Let's see what the limiting reactant is. Assuming NO is limit. 2.9761 mol / 2 * 5 = 7.44 mol So 2.9761 moles of NO will need 7.44 moles of hydrogen gas which is less than the amount of hydrogen we have. So NO is our limiting reactant. And since 2 moles of NO produces 2 moles of NH3, we will get the same number of moles of NH3 as moles of NO we have, so we'll have 2.9761 moles of NH3. To find the mass, just multiply by the molar mass. So 2.9761 mol * 17.03052 g/mol = 50.68453057 g Rounding to 3 significant figures gives 50.7 g.

Answer 2

Balanced equation: 
2 NO + 5 H2 ------> 2 NH3 + 2 H2O
 
2 moles NO react with 5 moles H2 to produce 2 moles NH3
 
Molar mass of NO = 30.00 g/mol 
86.3g NO = 86.3/30.00 = 2.877 moles of NO 

This will require: 2.877*5 / 2 = 7.192 moles of H2 

Molar mass of H2 = 2 g/mol 
25.6g H2 = 25.6/2 = 12.7 mol H2. 
You have excess H2 means the NO is limiting 

From the balanced equation: 
2 moles of NO will produce 2 moles of NH3 
2.877 moles of NO will produce 2.877 moles of NH3 

Molar mass NH3 = 17g/mol 
Mass NH3 produced = 2.877 * 17 = 48.91g 

Hence the yield is = 48.91 g ~ 49 g