The given condition is STP, under this condition, gas has a rule of 22.4 L per mole. And the given equation is already balanced. The ratio of mole number is the same as the ratio of the volume and is also the same as the ratio of coefficients. So the answer is 4.0 liters.
Answer:
3920kg of lime(CaO) 1568m^3 of CO2
Explanation:
I attached the explanation!
also for calculating the volume I assumed Standard Temperature and Pressure(STP).
Answer:
C20 H14 O2
Explanation:
Remark
This is a sample, which the question does not say and should. It is a fraction of 1 mole. So what you have to do is multiply the numbers given by x and equate it to 286.28
Equation
150,86* x + 8.86*x + 20.1*x = 286.28
179.8x = 286.28
x = 286.26/179.8
x = 1.592
Now multiply the given numbers by 1.592
150.86 * 1.592 = 240.58
8.85 * 1.592 = 14.1
20.1 * 1.592 = 32
Rounding you get
240/12 = 20
14.1/1 = 14
32/16 = 2
C20 H14 O2
Answer:
O2
Explanation:
for find the limiting reactant you must calculate the moles of the reactants from the amount that you have and from the MM:
MM FeS2 = 120n = 26.2g / 120g/mol = 0,218 mol
MM O2 = 32n = 5,44g/32g/mol = 0,17 mol
The limiting reactant is
O2
Answer:
The area of the block is 150 cm
Explanation:
