Question

a solution has a concentration of 0.845 m. how many mol of solute are in 745 ml of that solution? how many mol of
question 4 1 pts
propane combusts with oxygen according to the reaction: c3h8 + 5 o2 --> 3 co2 + 4 h2o
if 15.0 g of c3h8 and 50.0g of o2 are allowed to react, which is the limiting reactant?

Explanation:

Step1: Convert volume to liters

The volume is $745\ mL$. To convert it to liters, use the conversion factor $1\ L = 1000\ mL$. So $V=745\ mL\times\frac{1\ L}{1000\ mL}=0.745\ L$.

Step2: Use the molarity formula

The molarity formula is $M=\frac{n}{V}$, where $M$ is molarity, $n$ is the number of moles, and $V$ is volume in liters. Rearranging for $n$ gives $n = M\times V$. Given $M = 0.845\ M$ and $V=0.745\ L$, then $n=0.845\ mol/L\times0.745\ L = 0.630525\ mol\approx0.630\ mol$.

For the second - part:

Step1: Calculate moles of $C_3H_8$

The molar mass of $C_3H_8$ is $M_{C_3H_8}=(3\times12.01 + 8\times1.01)\ g/mol=44.11\ g/mol$. The number of moles of $C_3H_8$, $n_{C_3H_8}=\frac{m}{M}=\frac{15.0\ g}{44.11\ g/mol}=0.340\ mol$.

Step2: Calculate moles of $O_2$

The molar mass of $O_2$ is $M_{O_2}=2\times16.00\ g/mol = 32.00\ g/mol$. The number of moles of $O_2$, $n_{O_2}=\frac{m}{M}=\frac{50.0\ g}{32.00\ g/mol}=1.5625\ mol$.

Step3: Use the mole - ratio from the balanced equation

From the balanced equation $C_3H_8 + 5O_2
ightarrow3CO_2+4H_2O$, the mole - ratio of $C_3H_8$ to $O_2$ is $1:5$. For $0.340\ mol$ of $C_3H_8$, the moles of $O_2$ required is $n_{O_2\ required}=0.340\ mol\times5 = 1.70\ mol$. But we have only $1.5625\ mol$ of $O_2$. So $O_2$ is the limiting reactant.

Answer:

Question 3: 0.630
Question 4: O2