moles - mass calculation\nexample: what is the mass of 1.33 moles of ba3(po4)2?\nmass - moles…

Explanation:

Step1: Calculate molar - mass of $Ba_3(PO_4)_2$

The molar mass of $Ba$ is approximately $137.33\ g/mol$, $P$ is approximately $30.97\ g/mol$, and $O$ is approximately $16.00\ g/mol$. For $Ba_3(PO_4)_2$, the molar - mass $M$ is: [ \begin{align*} M&=3\times137.33 + 2\times(30.97+4\times16.00)\ &=411.99+2\times(30.97 + 64.00)\ &=411.99+2\times94.97\ &=411.99 + 189.94\ &=601.93\ g/mol \end{align*} ]

Step2: Calculate mass of $1.33$ moles of $Ba_3(PO_4)_2$

We use the formula $m = n\times M$, where $n = 1.33\ mol$ and $M = 601.93\ g/mol$. $m=1.33\times601.93\approx790.57\ g$

Step3: Calculate molar - mass of $CaCl_2$

The molar mass of $Ca$ is approximately $40.08\ g/mol$ and $Cl$ is approximately $35.45\ g/mol$. For $CaCl_2$, the molar - mass $M'$ is $M'=40.08 + 2\times35.45=40.08+70.90 = 110.98\ g/mol$

Step4: Calculate moles of $157.8\ g$ of $CaCl_2$

We use the formula $n=\frac{m}{M}$, where $m = 157.8\ g$ and $M'=110.98\ g/mol$. $n=\frac{157.8}{110.98}\approx1.42\ mol$

Answer:

The mass of $1.33$ moles of $Ba_3(PO_4)_2$ is approximately $790.57\ g$ and the number of moles in $157.8\ g$ of $CaCl_2$ is approximately $1.42\ mol$.