posted by Anonymous on .
How many grams of nitrogen are required to react with 2.79g of hydrogen to produce ammonia?
A) 25.8 g B) 13.0 g C) 78.2 g D) 38.7 g E) 77.4 g
I get the answer as 38.7 but it is meant to be B
I did 2.79g x 3molH2/2.016g/mol x 1molN2/3molH2 x 28.02g/mol/1moleN2
Whats the equation to form NH2?
N2 +3H2 -----> 2NH3
So, you need 3 moles of H2 gas for 1 mole of N2:
How many moles of hydrogen are present in 2.79 g of Hydrogen?
2.79 g of H2*(1 mole/2.016g)=1.384 moles of H2
1.384 moles of H2 requires how many moles of N2?
1.384 moles of H2*(1 mole of N2/3 moles of H2)=0.4613 moles of N2
0.4613 moles of N2*(28.014g/mole)=12.9 g of N2=13.0g of N2.
I do it in steps when explaining. Also, it is better for accounting reasons in order to find errors. But you could do it all in one step, and you should still get the same answer:
2.79 g of H2*(1 mol/2.016g)*(1 mol N2/3 mol of H2)*(28.014g/mole)= 13.0g of N2