Question

moles → mass
given the following, find the number of grams:

1. 4 moles of cu(cn)₂

1. 5.6 moles of c₆h₆

1. 21.3 moles of baco₃

1. 1.2 moles of (nh₄)₂po₃

Explanation:

Problem 9: 4 moles of $\ce{Cu(CN)_{2}}$

Step 1: Calculate molar mass of $\ce{Cu(CN)_{2}}$

Molar mass of $\ce{Cu}$: $63.55\ \text{g/mol}$, $\ce{C}$: $12.01\ \text{g/mol}$, $\ce{N}$: $14.01\ \text{g/mol}$.
For $\ce{Cu(CN)_{2}}$: $1\times63.55 + 2\times(12.01 + 14.01) = 63.55 + 2\times26.02 = 63.55 + 52.04 = 115.59\ \text{g/mol}$.

Step 2: Use $m = n\times M$ (mass = moles × molar mass)

$n = 4\ \text{mol}$, $M = 115.59\ \text{g/mol}$.
$m = 4\times115.59 = 462.36\ \text{g}$.

Step 1: Calculate molar mass of $\ce{C_{6}H_{6}}$

Molar mass of $\ce{C}$: $12.01\ \text{g/mol}$, $\ce{H}$: $1.008\ \text{g/mol}$.
For $\ce{C_{6}H_{6}}$: $6\times12.01 + 6\times1.008 = 72.06 + 6.048 = 78.108\ \text{g/mol}$.

Step 2: Use $m = n\times M$

$n = 5.6\ \text{mol}$, $M = 78.108\ \text{g/mol}$.
$m = 5.6\times78.108 \approx 437.40\ \text{g}$.

Step 1: Calculate molar mass of $\ce{BaCO_{3}}$

Molar mass of $\ce{Ba}$: $137.33\ \text{g/mol}$, $\ce{C}$: $12.01\ \text{g/mol}$, $\ce{O}$: $16.00\ \text{g/mol}$.
For $\ce{BaCO_{3}}$: $137.33 + 12.01 + 3\times16.00 = 137.33 + 12.01 + 48.00 = 197.34\ \text{g/mol}$.

Step 2: Use $m = n\times M$

$n = 21.3\ \text{mol}$, $M = 197.34\ \text{g/mol}$.
$m = 21.3\times197.34 \approx 4203.34\ \text{g}$ (or ~4.20×10³ g).

Answer:

$462.36\ \text{g}$ (or ~462 g, depending on sig figs)

Problem 10: 5.6 moles of $\ce{C_{6}H_{6}}$