question in the description. please help

Question

42g/mol C2H6   32g/mol O2 44g/mol CO2 18g/mol H2O

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

201.6 g C2H6 reacts with 400.0g O2 to make water..

1. What is the theoretical yield of H2O based on the above reactant amounts?
2. Which substance is the limiting reactant?
3. What is the Percent yield of H2O for this reaction based on only 240g H2O being produced?
4. What mass of CO2 was produced for the above conditions?
5. What mass of C2H6 did not react?

Answer 1

The molar mass of C2H6 is 30.08 g/mol, not 40!

1:

201.6/30.08/2=3.35 molrxn C2H6

400.0/32/7=1.78 molrxn O2

O2 is limiting

Use limiting reactant to find theoretical yield

use molar mass

400gO2/32.00g=12.5 mol O2

use coefficients

12.5 mol O2 (6mol H2O/7molO2)=10.7 mol H2O

use molar mass

10.7molH2O (18.01g/1mol)=193g H2O

2

O2 was limiting

3

%yield=(actual/theoretical)100

(240/193)100=124.4% yield

4

use molar mass

240g H2O/18.02=13.3 mol

use coefficients

13.3molH2O (4molCO2/6molH2O)=8.88 mol CO2

5

mass C2H6=201.6g

mass C2H6 needed from O2:

use molar mass

400/32=12.5 mol O2

use coefficients

12.5 (2molC2H6/7molO2)=3.57 mol C2H6

use molar mass

3.57 mol C2H6(30.08g/1mol)=107.4 g C2H6 reacts

excess=initial-amount that reacts

excess=201.6-107.4

=94.2 g