In: Chemistry
1) A bottling plant has 1.26521×105 bottles with a capacity of 355 mL , 1.08937×105 caps, and 4.8740×104 L of beverage.
-How many bottles can be filled and capped, how many bottles left over, how much beverage left over, how many caps left over?
2) Hydrogen gas reacts with nitrogen gas to produce ammonia
3H2(g)+N2(g)→2NH3(g)
-What is the theoretical yield in grams and percent yield for this reaction?
3) Methane (CH4 ) burns and reacts with oxygen gas to produce carbon dioxide and water
CH4(g)+2O2(g)→CO2(g)+2H2O(g)
-What is the mass of carbon dioxide, water, and oxygen from this complete combustion of 1.40*10^-3 g of methane?
Reviewing for test using online appliction, application keeps rejecting answers, some of these concepts I thought I understood well I am struggling with these problems, first one seems simple but I am missing something big thank you in advance.
1.
Given bottles with out cap = 126521
Caps = 108937
number of bottles without cap = 126521 - 108937 = 17584
i.e., we have lack of 17,584 caps for the given bottles.
each one bottlle has a capacity of 0.355lit
then 48,740 lit of volume is filled in (48,740 / 0.355) = 137295.7 bottles are filled with liquid
therefore
number of bottles filled & capped = 108937
then number of bottles empty without cap and with out fill = 17584
volume of liquid remained = 48740 - 38672 = 10,067.365lit
In the remaining 17584 (without cap) bottles we can fill a volume of 6242.32lit
therefore, the volume of liquid remains (lack of bottles) = 10,067.365 - 6242.32 = 3825.045 lit
2.
limiting reactant is the hydrogen
3 moles of Hydrogen produces 2 moles of ammonia
i.e., 3 * 2 = 6gr of Hydrogen produce 2 * 17 = 34 grams of ammonia.
3.
from the reaction
1 moles of methane combines with 2 moles of oxygen
ie.., 16gr of methane combines with 2 * 16 = 32grams of oxygen
then 1.40 * 10-3 grams of methane combines with (32 * 1.40 * 10-3) / 16 = 2.8 * 10-3grams of oxygen
ii.
from the reaction
16 gr of methane giving 44gr of carbondioxide
then 1.40 * 10-3 gr of methane giving (44 * 1.40 * 10-3) / 16 = 3.85 * 10-3gr of carbon dioxide
iii.
16 gr of methane gives 36 grams of water
then 1.40 * 10-3 gr of methane produce (36 * 1.40 * 10-3) / 16 = 3.15 * 10-3 grams of water