Answer to Question #257619 in General Chemistry for Saint

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.