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By definition there are
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If 5.563 grams of $CH_4$ reacted with 25.00 grams of oxygen, how much $CO_2$ , in grams, can be produced? $CH_4(g) + 2O2(g) -> CO_2(g) + 2H_2O(g)$?
$"Moles of methane"$ $=$ $(5.563*g)/(16.043*g*mol^-1)$ $=$ $0.3468*mol$. $"Moles of dioxygen"$ $=$ $(25.00*g)/(31.998*g*mol^-1)$ $=$ $0.7813*mol$. Clearly there is sufficient dioxygen to react with methane as per the stoichiometric equation: $CH_4(g) +...
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Given this reaction: $CH_4 + 2O_2 -> CO_2 + H_2O$, 16 g of $CH_4$ react with 64 g of $O_2$, producing 44 g of $CO_2$. How many grams of water are produced?
There are actually two methods of doing it - stoichimetry & also with the help of law of conservation of mass. First by : $16$grams of $CH_4$ means $1$mole of...
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A sample of dolomitic limestone containing only $CaCO_3$ and $MgCO_3$ was analyzed. When a 0.2800 gram sample of this limestone was decomposed by heating, 75.0 mL of $CO_2$ at 750 mmHG and 20°C were evolved. How many grams of $CO_2$ were produced?
We use the Ideal Gas equation to access the moles of carbon dioxide evolved: $n=(PV)/(RT)$ $=$ $((750*mm*Hg)/(760*mm*Hg*atm^-1)xx75.0xx10^-3*L)/(0.0821*L*atm*K^-1*mol^-1xx293*K)$ $=3.1xx10^-3mol$. $"Mass of carbon dioxide produced"=3.1xx10^-3molxx44.0*g*mol^-1=??g$ The key to doing this problem was...
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How many $CO_2$ molecules are in $1.70 * 10^-2$ mole of $CO_2$?
.......where $N_A="Avogadro's Number"$ $=$ $6.022xx10^23*mol^-1$. And so, $1.70xx10^-2*molxxN_A=1.70xx10^-2*cancel(mol)xx6.022xx10^23*cancel(mol^-1)$, approx. $1.0xx10^22$ $"carbon dioxide molecules"$. We get the answer as a number as required.
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Hm many grams of $CO_2$ would be formed from 30 grams of $C_7H_8$ in the following reaction $C_7H_8 + O_2 -> CO_2 + H_2O$?
A balanced equation representing the complete combustion of toluene is absolutely necessary: $C_7H_8 + 9O_2 rarr 7CO_2 + 4H_2O$ This tells us that for every mole of toluene, 7 moles...
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If 3.00 mol of $CaCO_3$ undergo decomposition from $CaO$ and $CO_2$ how many grams of $CO_2$ are produced?
We need the stoichiometric equation; $CaCO_3(s) + Delta rarr CaO(s) + CO_2(g)uarr$ You have to heat this fairly fiercely, but clearly, quantitative reaction indicates that 1 mol of carbon dioxide...
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For $CH_4 +2O_2 -> CO_2 + 2H_2O$, the molar mass of oxygen gas ($0_2$) is 32.00 g/mol. The molar mass of carbon dioxide ($CO_2$) is 44.01 g/mol. What mass of $CO_2$, in grams, will form when 8.94 g $O_2$ completely react?
The balanced chemical equation tells us unequivocally that 16 g of methane gas reacts with 32 g of dioxygen gas to give 44 g of carbon dioxide, and 36 g...
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How many grams of $CO_2$ are produced from 125 g of $O_2$ and excess $CH_4$ in the equation $CH_4 +2O_2-> CO_2 + 2H_2O$?
Here $CH_4$ being excess $O_2 $ will be consumed fully. We know from above equation 2 mole or 64 g $O_2$ produces 1 mole or 44 g $CO_2$ So 125...
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$CO_2$ gas, in the dry state, may be produced by heating calcium carbonate. $CaCO_3 (s)DeltaCaO(s) + CO_2(g)$. What volume of $CO_2$, collected dry at 55 C and a pressure of 774 torr, is produced by complete thermal decomposition of 10.0 g of $CaCO_3$?
$10gramsCaCO_3(s) => CaO(s) + (?)O_2(g)$ $"moles" CaCO_3 = ((10g)/(100(g/(mol))))$ = $0.10"mole"$ From reaction ratios => moles $CaCO_3(s)$ consumed = moles $O_2(g)$ produced = 0.10 mole $(O_2(g))$**** If a 'dry'...
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The concentration of $CO_2$ in a can of soda is approximately 0.045 M. If all of the $CO_2$ in a 12 ounce soda evolves from solution, how many liters would the $CO_2$ gas take up at STP?
The idea here is that you need to figure out how many moles of carbon dioxide, $"CO"_2$, are initially dissolved in the soda by using the volume and of...
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