Oxygen and hydrogen react explosively to form water. In one reaction, 6 g of hydrogen combines with oxygen to form 54 g of water. How much oxygen was used?

I'll show you two approaches to solving this problem, one really short and one relatively long.

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SHORT VERSION

The problem tells you that $"6 g"$ of hydrogen gas, $"H"_2$, react with an unknown mass of oxygen gas, $"O"_2$, to form $"54 g"$ of water.

As you know, the tells you that the in a chemical reaction the total mass of the reactants must be equal to the total mass of the products.

In your case, this can be written as

$overbrace(m_(H_2) + m_(O_2))^(color(blue)("total mass of reactants")) = overbrace(m_(H_2O))^(color(orange)("mass of product"))$

This means that the reaction must have consumed

$m_(O_2) = m_(H_2O) - m_(H_2)$

$m_(O_2) = "54 g" - "6 g" = color(green)("48 g O"_2)$

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LONG VERSION

You can get the same result by using a bit of . First, write a balance d chemical equation for this reaction

$color(purple)(2)"H"_text(2(g]) + "O"_text(2(g]) -> color(red)(2)"H"_2"O"_text((g])$

Notice that you have a $color(red)(2):1$ between water and oxygen gas. This means that the reaction will produce twice as many moles of water as you have moles of oxygen taking part in the reaction.

Use water's molar mass to determine how many moles of water were produced

$54 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "2.998 moles H"_2"O"$

This means that the reaction consumed

$2.998 color(red)(cancel(color(black)("moles H"_2"O"))) * "1 mole O"_2/(color(red)(2)color(red)(cancel(color(black)("moles H"_2"O")))) = "1.499 moles O"_2$

Finally, use oxygen gas' molar mass to find how many grams would contain this many moles

$1.499 color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = color(green)("48 g")$

Once again, the answer will be $"48 g"$ of oxygen gas.