Hi, Can someone help me with this question, please! It's due tonight!
Given the balanced equation:

3 CO + 7 H2 C3H8 + 3 H2O,

how much C3H8 will be produced if 45.8 g CO reacts with 87.3 g H2?

Respuesta :

Answer:

24.0 g C₃H₈

Explanation:

To find the mass of C₃H₈, you need to (1) convert grams CO/H₂ to moles CO/H₂ (via molar mass), then (2) convert moles CO/H₂ to moles C₃H₈ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles C₃H₈ to grams C₃H₈ (via molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 3 sig figs to reflect the sig figs in the given values.

Molar Mass (CO): 12.011 g/mol + 15.998 g/mol

Molar Mass (CO): 28.009 g/mol

Molar Mass (H₂): 2(1.008 g/mol)

Molar Mass (H₂): 2.016 g/mol

Molar Mass (C₃H₈): 3(12.011 g/mol) + 8(1.008 g/mol)

Molar Mass (CH): 44.097 g/mol

3 CO + 7 H₂ ----> 1 C₃H₈ + 3 H₂O
^           ^              ^

45.8 g CO           1 mole              1 mole C₃H₈          44.097 g
-----------------  x  ------------------  x  --------------------  x  ------------------  =
                           28.009 g           3 moles CO            1 mole

=  24.0 g CH

87.3 g H₂           1 mole           1 mole C₃H₈          44.097 g
----------------  x  ---------------  x  ---------------------  x  -----------------  =  
                          2.016 g           7 moles H₂             1 mole

=  273 g CH

It was necessary to find the mass of the products from both of the reactants because you did not know which one was the limiting reagent. The limiting reagent is the reactant which is completely used up first. Because CO produced the smaller amount of product, it must be the limiting reagent. Therefore, the actual amount of CH produced is 24.0 grams.

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