For the reaction 2 C4H1O(g) +13O2(g)= 8CO2(g) +10H2O(l), 300g of C4H10 is combusted in 1000g of O2?

Question

what is the limiting reagent??

how many grams of H2O is formed???

Answer 1

2 mol C4H10 requires 13 mol O2

300 g / 58.1 g/mol = 5.161 mol C4H10

1000 g / 32.00 = 31.25 mol O2

5.161 mol C4H10 requires (13/2)x5.161 = 33.55 mol O2

O2 is limiting

13 mol O2 --> 10 mol H2O

31.25 mol O2 --> 24.0 mol H2O (31.25x10/13 = 24.0)

24.0 mol H2O x 18.02 g/mol = 433 g H2O formed