How many grams of sodium bicarbonate are required to neutralize 10.0 ml of 0.902 M vinegar?

8.4g

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1.5g

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0.75g

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1.07g

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Solution

The correct option is B 0.75g
$C H_{3} C O O H + N a H C O_{3} \to C H_{3} C O O N a + H_{2} O + C O_{2}$
Equivalents of $C H_{3} C O O H = M o l a r i t y \times v o l u m e \times n_{f} = 0.902 \times 10 \times 10^{- 3}$
Equivalents of $N a H C O_{3}$ required $= M o l e s \times n_{f} = 0.902 \times 10 \times 10^{- 3}$
Moles of $N a H C O_{3} = 9.02 \times 10^{- 3}$
Weight of $N a H C O_{3} = 9.02 \times 10^{- 3} \times 84 = 0.75 g r a m s$

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