Answer:
109.7178g of H2O
Explanation:
First let us generate a balanced equation for the reaction. This is illustrated below:
2C3H8O + 9O2 —> 6CO2 + 8H2O
Next we will calculate the molar mass and masses of C3H8O and H20. This is illustrated below:
Molar Mass of C3H8O = (3x12.011) + (8x1.00794) + 15.9994 = 36.033 + 8.06352 + 15.9994 = 60.09592g/mol.
Mass of C3H8O from the balanced equation = 2 x 60.09592 = 120.19184g
Molar Mass of H2O = (2x1.00794) + 15.9994 = 2.01588 + 15.9994 = 18.01528g/mol
Mass of H2O from the balanced equation = 8 x 18.01528 = 144.12224g
From the equation,
120.19184g of C3H8O produced 144.12224g of H20.
Therefore, 91.5g of C3H8O will produce = (91.5 x 144.12224) /120.19184 = 109.7178g of H2O
<span><span>Number of Protons-19 </span><span>Number of Neutrons-20 </span><span>Number of Electrons-<span>19</span></span></span>
Answer:
44.8 L of O2 will react (option D)
Explanation:
Step 1: Data given
Number of moles of SO2 = 4.00 moles
STP = Pressure = 1 atm and temperature = 273 K
Step 2: The balanced equation
2 SO2(g) + O2(g) → 2 SO3(g)
Step 3: Calculate moles of O2
For 2 moles SO2, we need 1 mol O2 to produce 2 moles SO3
For 4.00 moles SO2 we need 4.00 / 2 = 2.00 moles O2
Step 4: Calculate volume of O2
For 1 mol we have a volume of 22.4 L
V = (n*R*T)/ p
V = (2.00 * 0.08206 * 273)/p
V = 44.8 L
For 2.00 moles we have a volume of 2*22.4 = 44.8 L
44.8 L of O2 will react (option D)
Decomposer, decomposers include fungous, bacteria, worms. They break down trash and bad things and concert it to soil.
