Not Set 13.3/100 Gradebook Print calculator Periodic Table Question 10 of 10 Sap

Question

Not Set 13.3/100 Gradebook Print calculator Periodic Table Question 10 of 10 Sapling Learning Nitrogen and hydrogen combine at high temperature, in the presence of a catalyst, to produce ammonia. Assume 0.290 mol of N2 and 0.911 mol of H2 are present initially Number After complete reaction, how many mol moles of ammonia are produced? Number How many moles of H2 remai mol Number How many moles of N2 remain? mol O hydrogen What is the limiting reactant? O nitrogen O Previous ® Give up & View solution 2 Check Answer O Next Exit

Explanation / Answer

From the balanced equation we can say that

1 mole of N2 requires 3 mole of H2

so 0.290 mole of N2 will requie

= 0.290 mole of N2 *(3 mole of H2/1 mole of N2)

= 0.87 mole of H2

but we have 0.911 mole of H2 so N2 is limiting reactanta

1 mole of N2 produces 2 mole of NH3 so

0.290 mole of N2 will produce 0.58 mole of ammonia

Moles of H2 remain = 0.911 mole - 0.87 mole = 0.041 mole of H2

moles of N2 remain = 0.00

Limiting reactant is nitrogen

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