When a $3.0*g$ mass of carbon is burnt in an $8.0*g$ mass of dioxygen, the carbon and the oxygen are stoichiometrically equivalent. Of course, the combustion reaction proceeds according to...
1 Answers 1 viewsHere 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...
1 Answers 1 views$"1 mol CO"_2=6.022xx10^(23) "molecules CO"_2"$ Use the equality above to determine the moles of carbon dioxide as shown below. $6.022xx10^22 cancel"molecules CO"_2xx(1 "mol CO"_2)/(6.022xx10^23 cancel"molecules CO"_2)="0.1000 mol CO"_2"$ rounded to...
1 Answers 1 viewsYour tool of choice here is Avogadro's constant, which essentially acts as the definition of a mole $color(blue)(ul(color(black)("1 mole CO"_2 = 6.022 * 10^(23)color(white)(.)"molecules CO"_2))) ->$ Avogadro's constant...
1 Answers 1 viewsNow $N_A=6.022xx10^23*"molecules"$. This specifies a molar quantity of stuff. By definition $N_A$ $""^12C$ ATOMS has a mass of $12.00*g$, and constitutes a molar quantity. The $"mole"$ is thus the link...
1 Answers 1 viewsFrom the equation above, 3 moles of carbon dioxide result from every mole of propane combusted. We had, $(7.51*g)/(44.1*g*mol^-1)$ $=$ $?? "moles of propane"$. So how many moles of...
1 Answers 1 views$CS_2(l) + 3O_2(g) rarr CO_2(g) + 2SO_2(g)$ The stoichiometric equation says that $1*mol$ $CS_2$, $76*g$, reacts with $3*mol$ dioxygen, $96.0*g$ to give $44*g$ $CO_2$, $1*mol$, and $2*mol$ $SO_2$, i.e. $128*g$....
1 Answers 1 views$BaCO_3(s) + 2HNO_3(aq) rarrBa(NO_3)_2(aq)+ H_2O(l) + CO_2(g)uarr$ Given the , ONE MOLE of carbon dioxide is evolved for each mole of barium carbonate. $"Moles of barium carbonate"=(211*g)/(197.34*g*mol^-1)=1.07*mol.$ Given the...
1 Answers 1 viewsEthene undergoes incomplete combustion to form carbon dioxide, carbon monoxide and water vapour. The balanced equation of this incomplete combustion reaction is as follows. $C_2H_4(g)+5/2O_2(g)->CO_2(g)+CO(g)+2H_2O(g)$ But as per question the...
1 Answers 1 views$"8.855 g/L"$ This is just asking you to calculate the new molar (i.e. molar solubility) at a different pressure, i.e. use the . (We'll end up converting...
1 Answers 1 views