How many grams of propane, C3H8, are needed to make 140 grams of water?

85.6 g

Explanation:

Step 1: Write the balanced combustion equation

C₃H₈ + 5 O₂ ⇒ 3 CO₂ + 4 H₂O

Step 2: Calculate the moles corresponding to 140 g of H₂O

The molar mass of H₂O is 18.02 g/mol.

140 g × 1 mol/18.02 g = 7.77 mol

Step 3: Calculate the moles of C₃H₈ needed to produce 7.77 moles of H₂O

The molar ratio of C₃H₈ to H₂O is 1:4. The moles of C₃H₈ needed are 1/4 × 7.77 mol = 1.94 mol.

Step 4: Calculate the mass corresponding to 1.94 moles of C₃H₈

The molar mass of C₃H₈ is 44.10 g/mol.

1.94 mol × 44.10 g/mol = 85.6 g

i think slash and

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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