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Home barbeques are fueled with propane gas (C_3H_8). What mass of carbon dioxide (in kg) forms upon the complete combustion of 18.9 propane (approximate contents of one 5-gallon tank)?
We address the combustion equation.. $H_3C-CH_2CH_3(g) + 5O_2(g) rarr 3CO_2(g) +4H_2O(l) + Delta$ And, assuming you are a septic, $"1 US Gallon"-=3.785*L$.. We combust a mass of ethane... $"5 US"...
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When propane, $C_3H_8$, is burned, carbon dioxide and water vapor are produced according to the following reaction: $C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O$. How much propane is burned if 160.0 g of O2 are used and 132 g of CO2 and 72.0 g of H2O are produced?
$"Moles of carbon dioxide:"$ $(132*g)/(44.0*g*mol^-1)=3*mol$ $"Moles of water:"$ $(72*g)/(18.01*g*mol^-1)=4*mol$ $"Moles of oxygen:"$ $(160*g)/(32.0*g*mol^-1)=5*mol$ Because there were $3$ $mol$ carbon dioxide produced, and stoichiometric water, one mole precisely of propane was...
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When glucose is consumed, it reacts with oxygen in the body to produce carbon dioxide, water, and energy. How many grams of carbon dioxide would be produced if 45 g of $C_6H_12O_6$ completely reacted with oxygen?
Here is the balanced equation: $C_6H_12O_6 + 6O_2 rarr 6 CO_2 + 6 H_2O$ Since the molar mass of glucose is 180 g, the 45 g given mass in this...
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While you cook on your gas grill, 3.7 grams of $CO_2$ are produced. How many moles of propane, $C_3H_8$ are used, using the equation $C_3H_8+5O_2 -> 3CO_2+4H_2O$?
Given mass of $CO_2$ = 3.7g Molar mass of $CO_2$ = 12+2(16)= 44g/mol No of moles of $CO_2$ produced = $("given mass")/("molar mass")$ = $3.7/44$ =0.084mol...
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In an experiment, 1 mol of propane is burned to form carbon dioxide and water: $C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O$. How many moles of oxygen are needed for the reaction?
$C_3H_8(g) + 5O_2(g)rarr3CO_2(g) + 4H_2O(g)$ The stoichiometric equation CLEARLY specifies that $"1 mol"$ ($44*g$) of propane reacts with $"5 moles"$ ($160*g$) of dioxygen to give $"3 moles"$ ($132*g$) of carbon...
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Propane and oxygen react according to the equation $C_3H_8(g) + 5O_2(g) -> 3CO_2(g) + 4H_2O(g)$. How many grams of water can be produced from the complete combustion of 2.0 moles of $C_3H_8(g)$?
Since the molar mass of water $M(H_2O)=2xx1+16=18g//mol$ 2 mol propane will generate $2xx4xx18=144g$ of water
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When glucose is consumed, it reacts with oxygen in the body to produce carbon dioxide, water, and energy. How many grams of carbon dioxide would be produced if45 g of $C_6H_12O_6$ completely reacted with oxygen?
First, find the balanced chemical formula: $C_6H_12O_6 + 6O_2 rarr 6CO_2 + 6H_2O$ Since glucose forms carbon dioxide in a 1:6 ratio, we know that one mole of glucose yields...
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If I start with 5 grams of $C_3H_8$, what is the theoretical yield of water in the equation $C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O$?
the molar mass of$C_3H_8=3xx12+8xx1=44gmmol^=1$ molar mass of water=$18gmmol^-1$ As per the equation 44 gm$C_3H_8$ gives $4xx18=72gm$ water 1gm $C_3H_8$ gives $4xx18=72/44gm$ water 5gm $C_3H_8$ gives...
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Propane, $C_3H_8$, reacts with oxygen to produce carbon dioxide and water. If 10.0 grams of oxygen reacts with an excess of propane, then how many moles of carbon dioxide were produced?
Given: Mass of $"O"_2$; chemical equation (understood) Find: Moles of $"CO"_2$ Strategy: The central part of any problem is to convert moles of something to moles of something else....
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The combustion of propane ($C_3H_8$) produces 248 of energy per mole of propane burned. How much heat energy will be released when 1 000 L of propane are burned at STP?
We can represent propane combustion by the following equation: $C_3H_8(g) + O_2(g)rarr3CO_2(g) + 4H_2O(l) + 2220* kJ$ I took the heat of combustion values from this . Your question does...
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