The number of moles of $N a C l$ required to react with excess of $A g N O_{3}$ to produce $5.56 g$ of $A g C l$ is:

(Molar mass of $A g C l$ is $143.43 {g mol}^{- 1}$ }

0.038

moles

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0.026

moles

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0.048

moles

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0.056

moles

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Solution

The correct option is A $0.038$ moles
$A g N O_{3} + N a C l \to A g C l + N a N O_{3}$

Molar Mass of $A g C l$ = $143.32 g$

Number of moles of $A g C l$ in $5.56 g$ of $A g C l$ = $\frac{Given mass of A g C l}{Molar Mass of A g C l}$
$= \frac{5.56}{143.32} = 0.038 mol$

1 mole of $N a C l$ produces 1 mole of $A g C l$

So, the number of moles of $N a C l$ that produces $0.038 mol$ of $A g C l$
= $0.038 mol$

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Similar questions

N a C l

A g N O_{3}

14.343 g

A g C l

A g C l

143.43 {g mol}^{- 1}

Nitrogen monoxide reacts with oxygen according to the reaction:

2NO (g) + O₂(g) → 2NO₂(g)

If 12 moles of nitrogen monoxide reacts with 10 moles of oxygen, how many moles of NO₂ will be formed?

24.088 \times 10^{23} S O_{2}

6.022 \times 10^{18} H_{3} P O_{4}

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