Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    6
    ReputationRep:
    could somebody please explain how to do this question?

    Calculate the mass of H2O required to react completely with 5.0 g of SiCl4:
    SiCl4 + 2H2O  SiO2 + 4HCl
    • TSR Support Team
    Online

    21
    ReputationRep:
    Right, well you've got the balanced equation, so you can see that for every mole of SiCl4 you need 2 moles of H2O for a reaction.

    So then, convert to moles:
    5.0g of SiCl4 in moles is: 5.0g / 169.9gmol-1 (169.9 being the molecular weight of silicon tetrachloride) which is 0.0294... moles.

    So then, if for every one mole of SiCl4 reacting, 2 moles of H2O reacts, then the number of moles of H2O in the reaction is double, so 0.0294 * 2 = 0.588 moles of H2O reacting.

    Then convert to mass. Moles = mass/molar mass therefore mass = moles * molar mass so 0.588 * 18.0 = 1.058g of water
    • Thread Starter
    Offline

    6
    ReputationRep:
    so you work out the number of moles in SiCl4,
    and because the ratio of SiCl4 to H20 is 1:2, then you times the number of moles in SiCl4 by 2
    and then to find the mass of h20 that reacts with SiCl4, you convert the answer by using mass = moles x Ar.

    riiiiiiiiiiight, i get it, I don't know why i get so stuck on these questions, they're quite simple. Thanks for your help!
    • Thread Starter
    Offline

    6
    ReputationRep:
    sorry to be a pain but I don't understand this question either


    When copper (II) nitrate is heated, it decomposes according to the following equation:
    2Cu(NO3)2(s)  2CuO(s) + 4NO2(g) + O2(g).
    When 20.0g of copper (II) nitrate is heated, what mass of copper (II) oxide would be produced? What mass of NO2 would be produced?
    • TSR Support Team
    Online

    21
    ReputationRep:
    It's rather similar to the question above, again it's a case of work out the moles of Copper (II) Nitrate.

    Then, you apply the ratios again, the ratio of Copper (II) Nitrate to Copper (II) Oxide is 2:2 or rather 1:1, so the number of moles of Copper (II) Oxide is the same as that of the Copper (II) Nitrate, while the Copper Nitrate: Nitrogen Dioxide ratio is 2:4 or 1:2, so twice the number of moles of Nitrogen Dioxide is produced as the number of Copper Nitrate that reacted.

    So again, work out the moles of Copper Nitrate, then in separate equations, work out the number of moles of Copper Oxide and Nitrogen Dioxide formed, and convert that to mass.
    • Thread Starter
    Offline

    6
    ReputationRep:
    hmm, I did this for copper oxide and got 42.1, but the answer is apparently 8.48g :/

    i found the Ar of copper (II) nitrate, then divided the mass by the Ar so that I found the moles,
    then cause the ratio is the same i went straight into multiplying the moles by the Ar of Copper (II) oxide to find the mass

    so i'm confused as to where i went wrong ?
    • TSR Support Team
    Online

    21
    ReputationRep:
    Right, for the copper oxide:

    You first work out the number of molecular weight of copper nitrate - which is 187.55 gmol-1
    Then, the moles, which would be 20.0/187.55 = 0.107 mols
    As it's 1:1 ratio, the number of moles is the same, so times it by the molecular weight of copper oxide which is 79.54 gmol-1
    79.54 * 0.107 = 8.48g

    I suspect you either got the Ar of one of the copper compounds wrong, or had a stray digit when doing the calculations, could you post your full work-up for me to check through and find the error?
    • Thread Starter
    Offline

    6
    ReputationRep:
    ah fair enough. thanks for all the help dude. much appreciated!
    Offline

    1
    ReputationRep:
    I'm sooooooo stuck on the h2o one :/
    HELP PLEASEE
    Offline

    14
    ReputationRep:
    We dont have the mass for H2O, so we have to use the molecular masses from the periodic table to calculate it:

    SiCl4 -> 2H2O
    28+(35.5*4) -> 2*(2+16)
    170g -> 36g
    -----------------------------------
    Therefore: (Basic maths/ratio)
    1g (of SiCl4) -> 36/170 (of 2H2O)
    1g -> 0.21176470588

    So,
    5g -> (0.21176470588*5) = 1.06g OF H2O
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Will you be richer or poorer than your parents?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.