Question

How many grams are in 0.005 moles of lead (IV) sulfate, Pb(SO­4)2 Group of answer choices 1.5 g 502.74 g 0.0045 g 1.996 g

Answer 1

Answer:

$m_{Pb(SO_4)_2}=1.996 g$

Explanation:

Hello.

In this case, given the formula of the lead (IV) sulfate, we can compute its molar mass as shown below:

$M_{Pb(SO_4)_2}=m_{Pb}+2*m_S+2*4*m_O\\\\M_{Pb(SO_4)_2}=207.2+2*32+2*4*16\\\\M_{Pb(SO_4)_2}=399.2g/mol$

Thus, by applying the following mole-mass relationship, we obtain the mass of 0.005 moles of lead (IV) sulfate:

$m_{Pb(SO_4)_2}=0.005mol*\frac{399.2g}{1mol} \\\\m_{Pb(SO_4)_2}=1.996 g$

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