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    3kg ammoia used in reaction 1, all of the NO produced was used to make NO2 gas in reaction2, calculate mass of NO2 formed from 3kg ammonia in reacyion 2, assuming an 80%yield.
    reaction 1: 4NH3 +502----->4NO +6H20
    reaction 2: 2NO+02-------->2NO
    I calculated moles of ammonia =3000/17 =176.47
    but for the mass in the mrkschme it sAYs multiply 176.47 by 46??????
    but why don't you divide by 2 because the molar ratios aren't even equal?

    3kg ammoia
    calculate the mass of NO2
    reaction 1: 4NH3 +502----->4NO +6H20
    reaction 2: 2NO+02-------->2NO2

    reaction 1
    number of moles of NH3 = 3000/17 = 176.47
    mole ratio(4:4 divide by 4) 1:1 so number of moles of NO = 176.47

    reaction 2
    all the NO from reaction 1 goes to reaction 2
    moles of NO = 176.47
    mole NO2 (2:2) 1:1 mole ratio = 176.47
    mass of NO2 = mass * mr = 176.47 * (14+16+16) = 176.47 * 47
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