Question

When 35.0g of Ba(NO3)2 is reacted with excess Na2SO4 29.8g of BaSO4 is recovered by the chemist. What is the theoretical yield of BaSO4?

Answer 1

Write down the reaction equation:

Ba(NO₃)₂ + Na₂SO₄ => BaSO₄ + 2 NaNO₃

From the equation it is clear that 1 mol of Ba(NO₃)₂ allows to obtain 1 mol of BaSO₄

Molar mass of Ba(NO₃)₂ is 261.3 g/mol

Molar mass of BaSO₄ is 233.4 g/mol

Find the theoretical yield of BaSO₄:

$m$_theor = $m_1*M_2/M_1 = 35.0 * 233.4 / 261.3 = 31.3 (g)$

Answer: 31.3 g -- the theoretical yield of BaSO₄.