CH4+2 O2-> CO2 + 2 H2O You burned 15.0 grams of methane (CH4) in an excess of oxygen and formed 23.8 grams of water. What was your percent yield, rounded to 3 sig-figs?
C= 12.001 g/mol
H= 1.008 g/mol
O= 16g/mol

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2

these nuts