A mixture of 4.50 g H2 and 23.71 g O2 reacts to form water. What substances are

Question

A mixture of 4.50 g H2 and 23.71 g O2 reacts to form water. What substances are present after the reaction is over? (Hint: Write a balanced chemical equation for the reaction first.) How many grams of water are formed?

Explanation / Answer

Balanced Chemical Equation: 2H2 + O2 --> 2H2O First, we solve for the moles of H2 from its molar mass: 4.50 g x 1 mol/2.02 g = 2.23 moles Next, we solve for the moles of O2 from its molar mass: 23.71 g x 1 mol/32.00 g = 0.741 moles Since O2 produces the fewer number of moles, it is the LR and therefore used to solve for the moles of H2O: 0.741 moles of O2 x 2 moles of H2O/1 mole of O2 = 1.48 moles Now, from the moles of H2O, we can solve for its mass from its molar mass: 1.48 moles x 18.02 g/mol = 26.7 g Hope this helps! :)

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