Answer to Question #268162 in General Chemistry for Elif

Question

415 ml of 0.275 M copper(II) nitrate solution is reacted to 25.8 g of sodium bicarbonate (NaHCO₃) to form copper(II) carbonate, sodium nitrate, water and carbon dioxide. If reaction took place with 75% yield, how many grams of copper(II) carbonate forms?

Answer 1

$2 NaHCO_3 + Cu(NO_3)_2 → 2 NaNO_3 + CuCO_3 + CO_2 + H_2O$

Moles of $Cu(NO_3)_2=0.275×0.415=0.114moles$

Moles of $NaHCO_3=\frac{25.8}{ 123.55}=0.209$

Moles of $CuCO_3=\frac{0.209}{2}=0.1045moles$

Mass of $CuCO_3=$ $123.55×0.1045=12.91grams$