Propane and oxygen react according to the equation #C_3H_8(g) + 5O_2(g) -> 3CO_2(g) + 4H_2O(g)#. How many grams of water can be produced from the complete combustion of 2.0 moles of #C_3H_8(g)#?
From the equation you will see that 1 mol of propane generates 4 mols of water.
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To determine the grams of water produced from the complete combustion of 2.0 moles of C3H8(g), you first need to find the mole ratio between C3H8 and H2O in the balanced chemical equation. The equation shows that for every mole of C3H8, 4 moles of H2O are produced. Therefore, with 2.0 moles of C3H8, you can produce:
2.0 moles C3H8 * (4 moles H2O / 1 mole C3H8) = 8.0 moles H2O
Next, you need to convert moles of H2O to grams using the molar mass of water (H2O), which is approximately 18.015 grams/mol.
8.0 moles H2O * (18.015 grams H2O / 1 mole H2O) ≈ 144.12 grams H2O
Therefore, approximately 144.12 grams of water can be produced from the complete combustion of 2.0 moles of C3H8(g).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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