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Bur asked in Science & MathematicsChemistry · 2 years ago

Gas Stoichiometry?

0.500 g of methanol () is combusted in a test tube with 0.600 g of oxygen gas. The test tube is covered

with a deflated balloon. What will the approximate volume of the balloon be after the reaction is

complete if the experiment is conducted at SATP?

2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(g)

1 Answer

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  • Dr W
    Lv 7
    2 years ago
    Favorite Answer

    2 CH3OH(l) + 3 O2(g) → 2 CO2(g) + 4 H2O(g)... .. <--- you sure water is in the (g) phase?

    if CH3OH is the LR

    .. 0.500g CH3OH.. .. 1 mol CH3OH... ... 6 mol gas

    ----- ---- ---- ---- --- x ---- ----- ---- ---- x ---- ---- ---- ---- --- = 0.04682 mol gas

    .. .. ... ... .1. ... ... .... 32.04g CH3OH.. ..2 mol CH3OH

    if O2 is the LR

    .. 0.60g O2... . 1 mol O2.. .6 mol gas

    ---- ---- ---- x ---- ----- --- x ---- ---- ----- = 0.0375 mol gas

    .. .. .. 1.. .. .. .32.00g O2.. ..3 mol O2

    since O2 gave less gas than methanol, O2 is the LR and 0.0375 moles of gas is formed

    ************

    PV = nRT

    .. V = nRT/P

    .. V = (0.0375mol) * (0.08206 Latm/molK) * (__K) / (__ atm)

    you get to figure out what STP means and plug in the appropriate values and calculate the results

    https://en.wikipedia.org/wiki/Standard_conditions_...

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