K:
m=155g
M=39g/mol
n = 155g / 39g/mol ≈ 3,97mol
KNO₃:
m=122g
M=101g/mol
n = 122g/101g/mol = 1,21mol
2K + 10KNO₃ ⇒ 6K₂O + N₂
2mol : 10mol
3,97mol : 1,21mol
limiting reagent
KNO₃ is limiting reagent
Answer:
<h3>
moles of carbon dioxide=13.95mol</h3>
Explanation:
First wrie down the balance chemical reaction:

Combustion reaction: The reacion in which hydrocarbon is burnt in the presence of oxygen gas and it releases heat and this reaction exothermic because heat of cumbustion is negative.
eg. burning of methane
By using unitry method,
From the above balanced reaction it is clearly that,
from 1 mole of propane 3 moles of carbon dioxide is formed
there fore,
from 4.65 mole of propane
moles of carbon dioxide will form
moles of carbon dioxide=13.95mol
Answer:
The estimated feed rate of logs is 14.3 logs/min.
Explanation:
The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.
That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.
Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.
To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.

The specific gravity of the wood chips is 0.494.
The average volume of a log is

The weight of one log is

To provide 3617 ton/day of wood chips, we need


The feed rate of logs is 14.3 logs/min.
<span>In order to calculate the surface are of the sheet of paper in square centimeters you must first convert the dimensions of the paper to centimeters. You can do this by multiplying 8.5 * 2.54 and 6.5 * 2.54. The paper measures 21.59cm by 16.51cm. To find the surface area you multiply the dimensions of the paper to equal 356.4509cm2</span>