i havent take it chemistry for a long time so i forgot how to do the caculation

Question

i havent take it chemistry for a long time so i forgot how to do the caculation of the formulate can someone help me and show me how to do it thank you calculate how many g of ammonia will be produce from 9.35 g ofnitrogen gas and excess hydrogen using the following equation N2(g) + 3 H2(g) --------------> 2NH3(g) i havent take it chemistry for a long time so i forgot how to do the caculation of the formulate can someone help me and show me how to do it thank you calculate how many g of ammonia will be produce from 9.35 g ofnitrogen gas and excess hydrogen using the following equation N2(g) + 3 H2(g) --------------> 2NH3(g)

Explanation / Answer

First calculate the moles of nitrogen gas you have:     moles N2 = mass / molar mass ofN2                    = 9.35 g / (14.01*2) g.mol-1                    = 0.3337 mol The equation is balanced already, and it tells us that 1 mole ofnitrogen gas (N2) react to form 2 moles of ammonia(NH3). Therefore we can calculate the moles of ammoniaproduced:     moles NH3 = 2*moles N2                      = 2*0.3337                      = 0.6674 mol Now do the reverse of the first calculation to find the mass ofammonia produced:     mass = moles * molar mass of NH3             = 0.6674 mol * (14.01 + 1.008*3) g.mol-1             = 11.4 g NH3

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