112g. of propane (C3 H8) is combusted mention the products write balanced chemical equation and calculate the mass of the carbon dioxide evolved in the reaction [H(A=1U), C(A=12H),O(A=16U)]
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The molar masses of propane and water are 44 g/mol and 18 g/mol respectively.
44.0 grams of propane corresponds to 1 mole of propane.
According to the balance chemical equation, C
3
H
8
(g)+5O
2
(g)→3CO
2
(g)+4H
2
O(l),
1 mole of propane gives 4 moles of water.
Thus, the combustion of 44.0 g (1 mole) of propane will give 4 moles of water.
Mass of water obtained =4×18=72.0 g.
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