Home
Class 11
CHEMISTRY
3.0 g of H(2) react with 29.0 g O(2) to ...

3.0 g of `H_(2)` react with 29.0 g `O_(2)` to yield `H_(2)O`
(i) What is the limiting reactant ?
(ii) Calculate the maximum amount of water that can be formed
(iii) Calculate the amount of one of the reactants which remains unreacted.

Text Solution

AI Generated Solution

To solve the problem step by step, we will follow these steps: ### Step 1: Write the balanced chemical equation. The balanced chemical equation for the reaction between hydrogen and oxygen to form water is: \[ 2H_2 + O_2 \rightarrow 2H_2O \] ### Step 2: Calculate the number of moles of each reactant. ...
Promotional Banner

Topper's Solved these Questions

  • MOLE CONCEPT

    DINESH PUBLICATION|Exercise Board Examinations|35 Videos

  • MOLE CONCEPT

    DINESH PUBLICATION|Exercise N.C.E.R.T (Short answer)|13 Videos

  • HYDROGEN

    DINESH PUBLICATION|Exercise LONG|1 Videos

  • NOMENCLATURE OF ORGANIC COMPOUNDS

    DINESH PUBLICATION|Exercise MCQs|13 Videos

Similar Questions

Explore conceptually related problems

3.0 g of H_(2) react with 30.0 g of O_(2) yield H_(2)O . (i) Which is the limiting reagent? (ii) Calculate the maximum amount of H_(2)O that can be formed. (iii) Calculate the amount of reactant left unreacted.

When 3.0 g of H_(2) react with 29.Og of O_(2) to form water, O_(2) is the limiting reactant. statement is true or false

Limiting reactant: Urea [(NH_(2))_(2) CO] used as ferlilzer as animal feed, and in polymer industry, is prepared by the reaction between ammonia and carbon dioxide: 2NH_(3)(g) + CO_(2) (g) rarr (NH_(2))_(2) CO(aq.) + H_(2) O(1) In one process , 637.2g of NH_(3) is allowed to react with 11.42g of CO_(2) (i) Which of the two reactants is the limiting reactant? (ii) Calculate the mass of (NH_(2))_(2) CO formed? (iii) How much of the excess reagent (in grams) is left at the end of the reaction? Strategy: (i) Since we cannot tell by inspection which of the two recantants is the limiting reacant, we have to procced by first converting their masses into number of moles. Take each reactnat in turn and ask how many moles of product (urea) would be obtained if each were completely consumed. The reactant that gives the smaller number of moles of producet is the limiting reactant. (ii) Convert the moles of product obtained to grams of product. (iii) From the moles of product, calculate to grams fo excess reactant needed int he reaction. Then subtract this qunitity from the grams of the reactant available to find the quanity of the excess reactant remaining.

6 g of H_(2) reacts with 14gN_(2) to form NH_(3) till the reaction completely consumes the limiting reagent.The amount of other reactant (in g ) left are .......