To solve this we assume that the hydrogen gas is an
ideal gas. Then, we can use the ideal gas equation which is expressed as PV =
nRT. At a constant pressure and number of moles of the gas the ratio T/V is
equal to some constant. At another set of condition of temperature, the
constant is still the same. Calculations are as follows:
T1 / V1 = T2 / V2
V2 = T2 x V1 / T1
V2 = (100 + 273.15) K x 2.50 L / (-196 + 273.15) K
<span>V2 = 12.09 L</span>
Therefore, the volume would increase to 12.09 L as the temperature is increased to 100 degrees Celsius.
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Answer:co2
Explanation:because of the oxygen levels
The correct answer is approximately 11.73 grams of sulfuric acid.
The theoretical yield of water from Al(OH)3 is lower than that of H₂SO₄. As a consequence, Al(OH)3 is the limiting reactant, H₂SO₄ is in excess.
The balanced equation is:
2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O
Each mole of Al(OH)3 corresponds to 3/2 moles of H₂SO₄. The molecular mass of Al(OH)3 is 78.003 g/mol. There are 15/78.003 = 0.19230 moles of Al(OH)3 in the five grams of Al(OH)3 available. Al(OH)3 is in limiting, which means that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
The molar mass of H₂SO₄ is 98.706 g/mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.706 = 28.289 g
40 grams of sulfuric acid is available, out of which 28.289 grams is consumed. The remaining 40-28.289 = 11.711 g is in excess, which is closest to the first option, that is, 11.73 grams of H₂SO₄.
Answer:
if i consider this reaction
Fe2O3+ 3CO---》2Fe+ 3CO2
so let's calculate first moles of Fe2O3 i.e. = 256/159.69= 1.6 moles
So the one moles of Fe2O3 is forming three moles of CO2
hence 1.6 moles will form 4.8 moles of CO2
one mole of CO2 is 44 g so 4.8 moles of Co2 is 44×4.8= 211.2 g
so the conclusion is 211.2 g of CO2 can be produced from 256 g Fe2O3!!
i d k it's right or wrong but i tried my best :)
