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Question

If 0.5 mole of $$BaCl_2$$ is mixed with 0.2 mol of $$Na_3PO_4$$, the maximum amount of $$Ba_3(PO_4)_3$$ that can be formed is:



A
0.7 mol
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B
0.5 mol
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C
0.2 mol
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D
0.1 mol
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Solution

The correct option is D 0.1 mol
$$BaCl_2+Na_3PO_4\rightarrow Ba_3(PO_4)_2+NaCl$$
Equivalents of $$BaCl_2=0.5\times n_f=0.5\times 2=1$$
Equivalents of $$Na_3PO_4=0.2\times n_f=0.5\times 3=0.6$$
Equivalents of $$Ba_3(PO_4)_2 formed=0.6=moles \times n_f$$
Moles of $$Ba_3(PO_4)_2=\dfrac{0.6}{6}=0.1\ moles$$

Chemistry

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