Consider the equation. mc016-1.jpg Which represents the correct mass-volume relationship at STP? ~ 32.00 g of O2 react with an excess of H2 to produce (2 x 22.4) L of H2O. ~ 2.00 g of H2 react with an excess of O2 to produce (2 x 22.4) L of H2O. ~ 32.00 g of O2 react with an excess of H2 to produce 22.4 L of H2O. ~ (2 x 2.00) g of H2 react with an excess of O2 to produce 22.4 L of H2O.
The picture is the jpg missing from the post.
The only way to do this is to convert the masses (given for each option) to moles, then use coefficients see how much you will produce. note, that at STP, 1 mole of any gas is 22.4 L
Then I believe it would be the first answer since 32g of O2 = 1 mole and for every 1 mole of O2 there are 2 moles of H20.
that is righttt
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