Question

In: Chemistry

1) A bottling plant has 1.26521×105 bottles with a capacity of 355 mL , 1.08937×105 caps,...

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.

Solutions

Expert Solution

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


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