What is the percent yield for a process in which 10.4g of CH3OH reacts and 10.1 g of CO2 is formed according to the following equation:
2CH3OH(I) + 302(g) → 2C02(g) + 4H2O(1)
A. 70.7%
B. 103%
C. 97.1%
D. 52.1%
E. 37.9%

A. 70.7%

Explanation:

In the first step lets compute the molar mass of CH₃OH and CO

Molar Mass of CH₃OH =  1(12.01 g/mol) + 4(1.008 g/mol) +1(16.00 g/mol)

                                     = 32.042 g/mol

Molar Mass of CO₂      = 1(12.01 g/mol) + 2(16.00 g/mol)  

                                     = 44.01 g/mol

Mass of only one reactant i.e. CH₃OH is given so  it must be the limiting reactant. Next, the theoretical yield is calculated directly as follows:

Given mass of CH₃OH is 10.4 g. So we have:

                                     10.4g CH₃OH

Convert grams of CH₃OH to moles of CH₃OH utilizing molar mass of CH₃OH as:

                          1 mol CH₃OH / 32.042 g CH₃OH

Convert CH₃OH to moles of CO₂ using mole ratio as:

                             2 mol CO₂ / 2 mol CH₃OH

Convert moles of  CO₂ to grams of  CO₂ utilizing molar mass of  CO₂ as:

                           44.01 g/mol CO₂ / 1 mol CO₂

Now calculating theoretical yield using above steps:

[ 10.4 g CH₃OH ]  [1 mol CH₃OH / 32.042 g CH₃OH ]  [2 mol CO₂ / 2 mol CH₃OH]  [44.01 g/mol CO₂ / 1 mol CO₂]

Multiplication is performed here. We are left with 10.4 and 44.01 g CO₂ from numerator terms in the above equation and 32.042 from denominator terms after cancellation process of above terms. So this equation becomes:

= ( 10.4 ) ( 44.01 ) g CO₂ / 32.042

= 457.704/32/042

=  14.28 g CO₂

Theoretical yield =  14.28 g CO₂  

Finally compute the percent yield for a process in which 10.4g of CH₃OH reacts and 10.1 g of CO₂ is formed:

percent yield = (actual yield / theoretical yield) x 100

As we have calculated theoretical yield which is 14.28 g CO₂ and actual yield is 10.1 g CO₂ So,

percent yield = (10.1 g CO₂ / 14.28 g CO₂) x 100%

                       = 0.707 x 100%

                       = 70.7 %

Hence option A 70.7% yield is the correct answer.

yes definetly because you are using energy

2b2t

explanation:

2b2t