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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 molecules of $CO_2$ are there in 55 grams of $CO_2$?
By definition there are $6.022xx10^23$ carbon dioxide molecules in a $44.0*g$ mass. How do I know this? Well because we use the given number to specify molar mass, the which...
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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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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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What are the moles OR masses associated with: (i) $15.27*g*"lithium metal"$; (ii) $2.40*mol*"sulfur"$; (iii) $0.0141*mol$ Ar; (iv) $88.0*mol*Mg$; (v) $2.29*g*"phosphorus"$; (vi) $11.9*mol*"chromium"$; (vii) $9.62*g;$ (viii) $237.6 mol As$?
This is the basis, and we use this quotient to approach all of the given problems. I trust that you do have a Periodic Table beside you, because you need...
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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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$CO_2$ 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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Which solution contains the biggest amount, in mol, of chloride ions?
A. 20 cm3 of 0.50 mol dm−3 NH4Cl
B. 60 cm3 of 0.20 mol dm−3 MgCl2
C. 70 cm3 of 0.30 mol dm−3 NaCl
D. 100 cm3 of 0.30 mol dm−3 ClCH2COOH
And to answer the question, we must calculate the of of each solution with respect to chloride ions. And $"number of moles"="concentration"xx"volume"$ For $(a):$ there are $0.020*dm^3xx0.50*mol*dm^-3=0.01*mol$. For $(b):$ $0.060*dm^3xx0.20*mol*dm^-3xx"2...
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